Engineering Economy 2012

Seventh Edition
ENGINEERING
ECONOMY

Seventh Edition
ENGINEERING
ECONOMY
Leland Blank , P. E.
Texas A & M University
American University of Sharjah, United Arab Emirates
Anthony Tarquin , P. E.
University of Texas at El Paso
TM

TM
ENGINEERING ECONOMY: SEVENTH EDITION
Published by McGraw-Hill, a business unit of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas, New
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Library of Congress Cataloging-in-Publication Data
Blank, Leland T.
Engineering economy / Leland Blank, Anthony Tarquin. — 7th ed.
p. cm.
Includes bibliographical references and index.
ISBN-13: 978-0-07-337630-1 (alk. paper)
ISBN-10: 0-07-337630-2
1. Engineering economy. I. Tarquin, Anthony J. II. Title.
TA177.4.B58 2012
658.15—dc22
2010052297
www.mhhe.com

This book is dedicated to Dr. Frank W. Sheppard, Jr. His lifelong
commitment to education, fair financial practices, international
outreach, and family values has been an inspiration to many—one
person at a time.

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CONTENTS
Preface to Seventh Edition xiii
LEARNING
STAGE 1
THE FUNDAMENTALS
Chapter 1 Foundations of Engineering Economy 2
1.1 Engineering Economics: Description and
Role in Decision Making 3
1.2 Performing an Engineering Economy Study 4
1.3 Professional Ethics and Economic Decisions 7
1.4 Interest Rate and Rate of Return 10
1.5 Terminology and Symbols 13
1.6 Cash Flows: Estimation and Diagramming 15
1.7 Economic Equivalence 19
1.8 Simple and Compound Interest 21
1.9 Minimum Attractive Rate of Return 25
1.10 Introduction to Spreadsheet Use 27
Chapter Summary 31
Problems 31
Additional Problems and FE Exam Review Questions 35
Case Study—Renewable Energy Sources for Electricity Generation 36
Case Study—Refrigerator Shells 37
Chapter 2 Factors: How Time and Interest Affect Money 38
PE Progressive Example—The Cement Factory Case 39
2.1 Single-Amount Factors (FP and PF ) 39
2.2 Uniform Series Present Worth Factor and Capital Recovery Factor (PA and AP) 43
2.3 Sinking Fund Factor and Uniform Series Compound Amount Factor (AF and FA) 46
2.4 Factor Values for Untabulated i or n Values 48
2.5 Arithmetic Gradient Factors (PG and AG) 50
2.6 Geometric Gradient Series Factors 58
2.7 Determining i or n for Known Cash Flow Values 61
Chapter Summary 64
Problems 64
Additional Problems and FE Exam Review Questions 69
Case Study—Time Marches On; So Does the Interest Rate 70
Chapter 3 Combining Factors and Spreadsheet Functions 72
3.1 Calculations for Uniform Series That Are Shifted 73
3.2 Calculations Involving Uniform Series and Randomly Placed Single Amounts 76
3.3 Calculations for Shifted Gradients 80
Chapter Summary 86
Problems 86
Additional Problems and FE Exam Review Questions 92
Case Study—Preserving Land for Public Use 93
Chapter 4 Nominal and Effective Interest Rates 94
PE Progressive Example—The Credit Card Offer Case 95
4.1 Nominal and Effective Interest Rate Statements 96
4.2 Effective Annual Interest Rates 99
4.3 Effective Interest Rates for Any Time Period 105
4.4 Equivalence Relations: Payment Period and Compounding Period 106
4.5 Equivalence Relations: Single Amounts with PP CP 107

viii
Contents
4.6 Equivalence Relations: Series with PP CP 109
4.7 Equivalence Relations: Single Amounts and Series with PP CP 112
4.8 Effective Interest Rate for Continuous Compounding 114
4.9 Interest Rates That Vary over Time 116
Chapter Summary 117
Problems 118
Additional Problems and FE Exam Review Questions 122
Case Study—Is Owning a Home a Net Gain or Net Loss over Time? 124
LEARNING
STAGE 2
BASIC ANALYSIS TOOLS
Chapter 5 Present Worth Analysis 128
PE Progressive Example—Water for Semiconductor Manufacturing Case 129
5.1 Formulating Alternatives 129
5.2 Present Worth Analysis of Equal-Life Alternatives 131
5.3 Present Worth Analysis of Different-Life Alternatives 133
5.4 Future Worth Analysis 137
5.5 Capitalized Cost Analysis 138
Chapter Summary 142
Problems 142
Additional Problems and FE Exam Review Questions 147
Case Study—Comparing Social Security Benefits 149
Chapter 6 Annual Worth Analysis 150
6.1 Advantages and Uses of Annual Worth Analysis 151
6.2 Calculation of Capital Recovery and AW Values 153
6.3 Evaluating Alternatives by Annual Worth Analysis 155
6.4 AW of a Permanent Investment 157
6.5 Life-Cycle Cost Analysis 160
Chapter Summary 164
Problems 164
Additional Problems and FE Exam Review Questions 169
Case Study—The Changing Scene of an Annual Worth Analysis 171
Chapter 7 Rate of Return Analysis: One Project 172
7.1 Interpretation of a Rate of Return Value 173
7.2 Rate of Return Calculation Using a PW or AW Relation 175
7.3 Special Considerations When Using the ROR Method 179
7.4 Multiple Rate of Return Values 180
7.5 Techniques to Remove Multiple Rates of Return 184
7.6 Rate of Return of a Bond Investment 190
Chapter Summary 193
Problems 193
Additional Problems and FE Exam Review Questions 198
Case Study—Developing and Selling an Innovative Idea 200
Chapter 8 Rate of Return Analysis: Multiple Alternatives 202
8.1 Why Incremental Analysis Is Necessary 203
8.2 Calculation of Incremental Cash Flows for ROR Analysis 203
8.3 Interpretation of Rate of Return on the Extra Investment 206
8.4 Rate of Return Evaluation Using PW: Incremental and Breakeven 207
8.5 Rate of Return Evaluation Using AW 213
8.6 Incremental ROR Analysis of Multiple Alternatives 214

Contents ix
8.7 All-in-One Spreadsheet Analysis (Optional) 218
Chapter Summary 219
Problems 220
Additional Problems and FE Exam Review Questions 225
Case Study—ROR Analysis with Estimated Lives That Vary 226
Case Study—How a New Engineering Graduate Can Help His Father 227
Chapter 9 Benefit/Cost Analysis and Public Sector Economics 228
PE Progressive Example—Water Treatment Facility #3 Case 229
9.1 Public Sector Projects 230
9.2 Benefit/Cost Analysis of a Single Project 235
9.3 Alternative Selection Using Incremental B/C Analysis 238
9.4 Incremental B/C Analysis of Multiple, Mutually Exclusive Alternatives 242
9.5 Service Sector Projects and Cost-Effectiveness Analysis 246
9.6 Ethical Considerations in the Public Sector 250
Chapter Summary 251
Problems 252
Additional Problems and FE Exam Review Questions 258
Case Study—Comparing B/C Analysis and CEA of Traffic Accident Reduction 259
LEARNING
STAGE 2
EPILOGUE: SELECTING THE BASIC ANALYSIS TOOL
LEARNING
STAGE 3
MAKING BETTER DECISIONS
Chapter 10 Project Financing and Noneconomic Attributes 266
10.1 MARR Relative to the Cost of Capital 267
10.2 Debt-Equity Mix and Weighted Average Cost of Capital 269
10.3 Determination of the Cost of Debt Capital 271
10.4 Determination of the Cost of Equity Capital and the MARR 273
10.5 Effect of Debt-Equity Mix on Investment Risk 275
10.6 Multiple Attribute Analysis: Identification and Importance of Each Attribute 278
10.7 Evaluation Measure for Multiple Attributes 282
Chapter Summary 283
Problems 284
Additional Problems and FE Exam Review Questions 289
Case Study—Which Is Better—Debt or Equity Financing? 290
Chapter 11 Replacement and Retention Decisions 292
PE Progressive Example—Keep or Replace the Kiln Case 293
11.1 Basics of a Replacement Study 294
11.2 Economic Service Life 296
11.3 Performing a Replacement Study 302
11.4 Additional Considerations in a Replacement Study 306
11.5 Replacement Study over a Specified Study Period 307
11.6 Replacement Value 312
Chapter Summary 312
Problems 313
Additional Problems and FE Exam Review Questions 319
Case Study—Will the Correct ESL Please Stand? 321

x
Contents
Chapter 12 Independent Projects with Budget Limitation 322
12.1 An Overview of Capital Rationing among Projects 323
12.2 Capital Rationing Using PW Analysis of Equal-Life Projects 325
12.3 Capital Rationing Using PW Analysis of Unequal-Life Projects 327
12.4 Capital Budgeting Problem Formulation Using Linear Programming 329
12.5 Additional Project Ranking Measures 332
Chapter Summary 334
Problems 334
Additional Problems and FE Exam Review Questions 338
Chapter 13 Breakeven and Payback Analysis 340
13.1 Breakeven Analysis for a Single Project 341
13.2 Breakeven Analysis Between Two Alternatives 345
13.3 Payback Analysis 348
13.4 More Breakeven and Payback Analysis on Spreadsheets 352
Chapter Summary 355
Problems 355
Additional Problems and FE Exam Review Questions 361
Case Study—Water Treatment Plant Process Costs 363
LEARNING
STAGE 4
ROUNDING OUT THE STUDY
Chapter 14 Effects of Inflation 366
14.1 Understanding the Impact of Inflation 367
14.2 Present Worth Calculations Adjusted for Inflation 369
14.3 Future Worth Calculations Adjusted for Inflation 374
14.4 Capital Recovery Calculations Adjusted for Inflation 377
Chapter Summary 378
Problems 379
Additional Problems and FE Exam Review Questions 384
Case Study—Inflation versus Stock and Bond Investments 385
Chapter 15 Cost Estimation and Indirect Cost Allocation 386
15.1 Understanding How Cost Estimation Is Accomplished 387
15.2 Unit Method 390
15.3 Cost Indexes 391
15.4 Cost-Estimating Relationships: Cost-Capacity Equations 394
15.5 Cost-Estimating Relationships: Factor Method 395
15.6 Traditional Indirect Cost Rates and Allocation 397
15.7 Activity-Based Costing (ABC) for Indirect Costs 401
15.8 Making Estimates and Maintaining Ethical Practices 403
Chapter Summary 404
Problems 404
Additional Problems and FE Exam Review Questions 410
Case Study—Indirect Cost Analysis of Medical Equipment Manufacturing Costs 411
Case Study—Deceptive Acts Can Get You in Trouble 412
Chapter 16 Depreciation Methods 414
16.1 Depreciation Terminology 415
16.2 Straight Line (SL) Depreciation 418
16.3 Declining Balance (DB) and Double Declining Balance (DDB) Depreciation 419
16.4 Modified Accelerated Cost Recovery System (MACRS) 422
16.5 Determining the MACRS Recovery Period 426

Contents xi
16.6 Depletion Methods 427
Chapter Summary 429
Appendix 430
16A.1 Sum-of-Years-Digits (SYD) and Unit-of-Production (UOP) Depreciation 430
16A.2 Switching between Depreciation Methods 432
16A.3 Determination of MACRS Rates 435
Problems 438
Additional Problems and FE Exam Review Questions 442
Appendix Problems 443
Chapter 17 After-Tax Economic Analysis 444
17.1 Income Tax Terminology and Basic Relations 445
17.2 Calculation of Cash Flow after Taxes 448
17.3 Effect on Taxes of Different Depreciation Methods and Recovery Periods 450
17.4 Depreciation Recapture and Capital Gains (Losses) 453
17.5 After-Tax Evaluation 456
17.6 After-Tax Replacement Study 462
17.7 After-Tax Value-Added Analysis 465
17.8 After-Tax Analysis for International Projects 468
17.9 Value-Added Tax 470
Chapter Summary 472
Problems 473
Additional Problems and FE Exam Review Questions 481
Case Study—After-Tax Analysis for Business Expansion 482
Chapter 18 Sensitivity Analysis and Staged Decisions 484
18.1 Determining Sensitivity to Parameter Variation 485
18.2 Sensitivity Analysis Using Three Estimates 490
18.3 Estimate Variability and the Expected Value 491
18.4 Expected Value Computations for Alternatives 492
18.5 Staged Evaluation of Alternatives Using a Decision Tree 494
18.6 Real Options in Engineering Economics 498
Chapter Summary 503
Problems 503
Additional Problems and FE Exam Review Questions 509
Case Study—Sensitivity to the Economic Environment 510
Case Study—Sensitivity Analysis of Public Sector Projects—Water Supply Plans 511
Chapter 19 More on Variation and Decision Making under Risk 514
19.1 Interpretation of Certainty, Risk, and Uncertainty 515
19.2 Elements Important to Decision Making under Risk 518
19.3 Random Samples 523
19.4 Expected Value and Standard Deviation 526
19.5 Monte Carlo Sampling and Simulation Analysis 533
Chapter Summary 540
Problems 540
Additional Problems and FE Exam Review Questions 543
Case Study—Using Simulation and Three-Estimate Sensitivity Analysis 544
Appendix A Using Spreadsheets and Microsoft Excel © 547
A.1 Introduction to Using Excel 547
A.2 Organization (Layout) of the Spreadsheet 549
A.3 Excel Functions Important to Engineering Economy 550
A.4 Goal Seek—A Tool for Breakeven and Sensitivity Analysis 558
A.5 Solver—An Optimizing Tool for Capital Budgeting, Breakeven, and Sensitivity Analysis 559
A.6 Error Messages 560

xii
Contents
Appendix B Basics of Accounting Reports and Business Ratios 561
B.1 The Balance Sheet 561
B.2 Income Statement and Cost of Goods Sold Statement 562
B.3 Business Ratios 563
Appendix C Code of Ethics for Engineers 566
Appendix D Alternate Methods for Equivalence Calculations 569
D.1 Using Programmable Calculators 569
D.2 Using the Summation of a Geometric Series 570
Appendix E Glossary of Concepts and Terms 573
E.1 Important Concepts and Guidelines 573
E.2 Symbols and Terms 576
Reference Materials 579
Factor Tables 581
Photo Credits 610
Index 611

PREFACE TO SEVENTH EDITION
This edition includes the time-tested approach and topics of previous editions and introduces significantly
new print and electronic features useful to learning about and successfully applying the exciting
field of engineering economics. Money makes a huge difference in the life of a corporation, an
individual, and a government. Learning to understand, analyze, and manage the money side of any
project is vital to its success. To be professionally successful, every engineer must be able to deal with
the time value of money, economic facts, inflation, cost estimation, tax considerations, as well as
spreadsheet and calculator use. This book is a great help to the learner and the instructor in accomplishing
these goals by using easy-to-understand language, simple graphics, and online features.
What's New and What's Best
This seventh edition is full of new information and features. Plus the supporting online materials
are new and updated to enhance the teaching and learning experience.
New topics:
• Ethics and the economics of engineering
• Service sector projects and their evaluation
• Real options development and analysis
• Value-added taxes and how they work
• Multiple rates of return and ways to eliminate them using spreadsheets
• No tabulated factors needed for equivalence computations (Appendix D)
New features in print and online:
• Totally new design to highlight important terms, concepts, and decision guidelines
• Progressive examples that continue throughout a chapter
• Downloadable online presentations featuring voice-over slides and animation
• Vital concepts and guidelines identified in margins; brief descriptions available (Appendix E)
• Fresh spreadsheet displays with on-image comments and function details
• Case studies (21 of them) ranging in topics from ethics to energy to simulation
Retained features:
• Many end-of-chapter problems (over 90% are new or revised)
• Easy-to-read language to enhance understanding in a variety of course environments
• Fundamentals of Engineering (FE) Exam review questions that double as additional or
review problems for quizzes and tests
• Hand and spreadsheet solutions presented for many examples
• Flexible chapter ordering after fundamental topics are understood
• Complete solutions manual available online (with access approval for instructors)
How to Use This Text
This textbook is best suited for a one-semester or one-quarter undergraduate course. Students
should be at the sophomore level or above with a basic understanding of engineering concepts
and terminology. A course in calculus is not necessary; however, knowledge of the concepts in
advanced mathematics and elementary probability will make the topics more meaningful.
Practitioners and professional engineers who need a refresher in economic analysis and cost
estimation will find this book very useful as a reference document as well as a learning medium.
Chapter Organization and Coverage Options
The textbook contains 19 chapters arranged into four learning stages, as indicated in the fl owchart
on the next page, and five appendices. Each chapter starts with a statement of purpose and a specific
learning outcome (ABET style) for each section. Chapters include a summary, numerous

xiv
Preface to Seventh Edition
CHAPTERS IN EACH LEARNING STAGE
Composition by level
Chapter 1
Foundations of
Engineering Economy
Learning
Stage 1:
The
Fundamentals
Chapter 2
Factors: How Time and
Interest Affect Money
Chapter 3
Combining Factors and
Spreadsheet Functions
Chapter 4
Nominal and Effective
Interest Rates
Learning
Stage 2:
Basic
Analysis
Tools
Chapter 5
Present Worth
Analysis
Chapter 6
Annual Worth
Analysis
Chapter 7
Rate of Return
Analysis:
One Project
Chapter 8
Rate of Return
Analysis: Multiple
Alternatives
Chapter 9
Benefit/Cost Analysis
and Public Sector
Economics
Learning Stage 2 Epilogue
Selecting the Basic
Analysis Tool
Learning
Stage 3:
Making
Better
Decisions
Chapter 10
Project Financing and
Noneconomic Attributes
Chapter 11
Replacement and
Retention Decisions
Chapter 12
Independent Projects
with Budget Limitation
Chapter 13
Breakeven and
Payback Analysis
Learning
Stage 4:
Rounding
Out the
Study
Chapter 14
Effects of
Inflation
Chapter 15
Cost Estimation and
Indirect Cost Allocation
Chapter 16
Depreciation
Methods
Chapter 17
After-Tax Economic
Analysis
Chapter 18
Sensitivity Analysis
and Staged Decisions
Chapter 19
More on Variation
and Decision Making
under Risk
end-of-chapter problems (essay and numerical), multiple-choice problems useful for course review
and FE Exam preparation, and a case study.
The appendices are important elements of learning for this text:
Appendix A Spreadsheet layout and functions (Excel is featured)
Appendix B Accounting reports and business ratios
Appendix C Code of Ethics for Engineers (from NSPE)
Appendix D Equivalence computations using calculators and geometric series; no tables
Appendix E Concepts, guidelines, terms, and symbols for engineering economics
There is considerable flexibility in the sequencing of topics and chapters once the first six
chapters are covered, as shown in the progression graphic on the next page. If the course is designed
to emphasize sensitivity and risk analysis, Chapters 18 and 19 can be covered immediately

Chapter Organization and Coverage Options xv
CHAPTER AND TOPIC PROGRESSION OPTIONS
Topics may be introduced at the point indicated or any point thereafter
(Alternative entry points are indicated by )
Numerical progression
through chapters
Inflation
Cost
Estimation
Taxes and
Depreciation
Sensitivity, Staged
Decisions, and Risk
1. Foundations
2. Factors
3. More Factors
4. Effective i
5. Present Worth
6. Annual Worth
7. Rate of Return
8. More ROR
9. Benefit/Cost
10. Financing and
Noneconomic Attributes
11. Replacement
12. Capital Budgeting
13. Breakeven and
Payback
14. Inflation
15. Estimation
16. Depreciation
17. After-Tax
18. Sensitivity, Decision
Trees, and Real Options
19. Risk and Simulation
after Learning Stage 2 (Chapter 9) is completed. If depreciation and tax emphasis are vitally
important to the goals of the course, Chapters 16 and 17 can be covered once Chapter 6 (annual
worth) is completed. The progression graphic can help in the design of the course content and
topic ordering.

SAMPLE OF RESOURCES FOR
LEARNING OUTCOMES
Each chapter begins with a purpose, list
of topics, and learning outcomes
(ABET style) for each corresponding
section. This behavioral-based
approach sensitizes the reader to what
is ahead, leading to improved
understanding and learning.
LEARNING OUTCOMES
Purpose: Use multiple factors and spreadsheet functions to find equivalent amounts for cash flows that have nonstandard
placement.
SECTION TOPIC LEARNING OUTCOME
3.1 Shifted series • Determine the P , F or A values of a series
starting at a time other than period 1.
3.2 Shifted series and single cash
flows
• Determine the P , F , or A values of a shifted series
and randomly placed single cash flows.
3.3 Shifted gradients • Make equivalence calculations for shifted
arithmetic or geometric gradient series that
increase or decrease in size of cash flows.
Time value of money
It is a well-known fact that money makes money. The time value of money explains the change
in the amount of money over time for funds that are owned (invested) or owed (borrowed).
This is the most important concept in engineering economy.
CONCEPTS AND GUIDELINES
To highlight the fundamental building
blocks of the course, a checkmark and title
in the margin call attention to particularly
important concepts and decision-making
guidelines. Appendix E includes a brief
description of each fundamental concept.
IN-CHAPTER EXAMPLES
Numerous in-chapter examples
throughout the book reinforce the
basic concepts and make
understanding easier. Where
appropriate, the example is solved
using separately marked hand and
spreadsheet solutions.
EXAMPLE 4.6
A dot-com company plans to place money in a new venture capital fund that currently returns
18% per year, compounded daily. What effective rate is this ( a ) yearly and ( b ) semiannually?
Solution
( a) Use Equation [4.7], with r 0.18 and m 365.
Effective i % per year ( 1 —— 0.18
365 ) 365 1 19.716%
(b) Here r 0.09 per 6 months and m 182 days.
Effective i % per 6 months ( 1 —— 0.09
182 ) 182 1 9.415%
PROGRESSIVE EXAMPLES
Water for Semiconductor Manufacturing
Case: The worldwide contribution of
semiconductor sales is about $250 billion
per year, or about 10% of the world’s
GDP (gross domestic product). This industry
produces the microchips used in many
of the communication, entertainment,
transportation, and computing devices
we use every day. Depending upon the
type and size of fabrication plant (fab),
the need for ultrapure water (UPW) to
manufacture these tiny integrated circuits
is high, ranging from 500 to 2000 gpm
(gallons per minute). Ultrapure water is
obtained by special processes that commonly
include reverse osmosis deionizing
resin bed technologies. Potable water
obtained from purifying seawater or
brackish groundwater may cost from
$2 to $3 per 1000 gallons, but to obtain
UPW on-site for semiconductor manufacturing
may cost an additional $1 to $3 per
1000 gallons.
A fab costs upward of $2.5 billion to
construct, with approximately 1% of this
total, or $25 million, required to provide
the ultrapure water needed, including
the necessary wastewater and recycling
equipment.
A newcomer to the industry, Angular
Enterprises, has estimated the cost profiles
for two options to supply its anticipated
fab with water. It is fortunate to
PE
have the option of desalinated seawater
or purified groundwater sources in the
location chosen for its new fab. The initial
cost estimates for the UPW system are
given below.
Source
Equipment first
cost, $M
Seawater
(S)
20
Groundwater
(G)
22
AOC, $M per year 0.5 0.3
Salvage value, % of
first cost
Cost of UPW, $ per
1000 gallons
5 10
4 5
Angular has made some initial estimates
for the UPW system.
Life of UPW equipment 10 years
UPW needs
Operating time
1500 gpm
16 hours per
day for 250 days
per year
This case is used in the following topics
(Sections) and problems of this chapter:
PW analysis of equal-life alternatives
(Section 5.2)
PW analysis of different-life alternatives
(Section 5.3)
Capitalized cost analysis (Section 5.5)
Problems 5.20 and 5.34
bla76302_ch04_094-126.indd 106
Several chapters include a progressive
example—a more detailed problem statement
introduced at the beginning of the chapter and
expanded upon throughout the chapter in
specially marked examples. This approach
illustrates different techniques and some
increasingly complex aspects of a real-world
problem.
12/22/10 8:24 PM

INSTRUCTORS AND STUDENTS
Contents xvii
bla76302_ch08_202-227.indd 212 12/11/10 6:52 PM
ONLINE PRESENTATIONS
An icon in the margin indicates the
availability of an animated voice-over slide
presentation that summarizes the material in
the section and provides a brief example for
learners who need a review or prefer videobased
materials. Presentations are keyed to
the sections of the text.
Breakeven ROR 17%
EXAMPLE 13.8
Breakeven
Chris and her father just purchased Incremental a ROR small office 17% building for $160,000 that is in need of a lot
of repairs, but is located in a prime commercial area of the city. The estimated costs each year
for repairs, insurance, etc. are $18,000 the first year, increasing by $1000 per year thereafter.
Figure At an 8–6 expected 8% per year return, use spreadsheet analysis to determine the payback period
PW if versus the building i graph and is (a) PW kept versus for incremental 2 years and i graph, sold for Example $290,000 8.4 . sometime beyond year 2 or (b) kept
for 3 years and sold for $370,000 sometime beyond 3 years.
Solution
Figure 13–11 shows the annual costs (column B) and the sales prices if the building is kept 2
or 3 years (columns C and E, respectively). The NPV function is applied (columns D and F) to
determine when the PW changes sign from plus to minus. These results bracket the payback
period for each retention Figure period 7–12 and sales price. When PW 0, the 8% return is exceeded.
(a) The 8% return payback Spreadsheet period is application between 3 and of 4 ROIC years method (column using D). If the Goal building Seek, is Example sold 7.6 .
after exactly 3 years for $290,000, the payback period was not exceeded; but after 4 years
it is exceeded.
(b) At a sales price of $370,000, the 8% return payback period is between 5 and 6 years (column
F). If the building is sold after 4 or 5 years, the payback is not exceeded; however, a
sale after 6 years is beyond the 8%-return payback period.
bla76302_ch07_172-201.indd 189
MARR
NPV(8%,$B$4:B7)+$B$3 PV(8%,A7,,290000)
If kept 2 years and
sold, payback is
between 3 and 4
If kept 3 years and
sold, payback is
between 5 and 6
MARR
Filter 1 ROR 25%
Filter 2 ROR 23%
3.1 Calculations for Uniform Series That Are Shifted
When a uniform series begins at a time other than at the end of period 1, it is called a shifted
series. In this case several methods can be used to find the equivalent present worth P . For
example, P of the uniform series shown in Figure 3–1 could be determined by any of the
following methods:
• Use the P F factor to find the present worth of each disbursement at year 0 and add them.
• Use the F P factor to find the future worth of each disbursement in year 13, add them, and
then find the present worth of the total, using P F ( P F , i ,13).
• Use the F A factor to find the future amount F A ( F A , i ,10), and then compute the present
worth, using P F ( P F , i ,13).
• Use the P A factor to compute the “present worth” P 3 A ( P A , i ,10) (which will be located
in year 3, not year 0), and then find the present worth in year 0 by using the ( P F , i ,3) factor.
12/11/10 4:32 PM
SPREADSHEETS
The text integrates spreadsheets to show
how easy they are to use in solving virtually
any type of engineering economic analysis
problem. Cell tags or full cells detail
built-in functions and relations developed
to solve a specific problem.
Figure 13–11
Payback period analysis, Example 13.8
bla76302_ch13_340-364.indd 354
12/17/10 1:02 PM
bla76302_ch03_072-093.indd 73
12/7/10 7:26 AM
FE EXAM AND COURSE
REVIEWS
Each chapter concludes with several
multiple-choice, FE Exam–style
problems that provide a simplified
review of chapter material. Additionally,
these problems cover topics for test
reviews and homework assignments.
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
8.38 When conducting a rate of return (ROR) analysis
involving multiple mutually exclusive alternatives,
the first step is to:
(a) Rank the alternatives according to decreasing
initial investment cost
(b) Rank the alternatives according to increasing
initial investment cost
(c) Calculate the present worth of each alternative
using the MARR
(d) Find the LCM between all of the alternatives
8.39 In comparing mutually exclusive alternatives by
the ROR method, you should:
(a) Find the ROR of each alternative and pick
the one with the highest ROR
(b) S l h l i h i l
8.43 For these alternatives, the sum of the incremental
cash flows is:
Year A B
0 10,000 14,000
1 2,500 4,000
2 2,500 4,000
3 2,500 4,000
4 2,500 4,000
5 2,500 4,000
(a) $2500
(b) $3500
(c) $6000
(d) $8000

CASE STUDY
RENEWABLE ENERGY SOURCES FOR ELECTRICITY GENERATION
Background
Pedernales Electric Cooperative (PEC) is the largest
member-owned electric co-op in the United States with over
232,000 meters in 12 Central Texas counties. PEC has a capacity
of approximately 1300 MW (megawatts) of power, of
which 277 MW, or about 21%, is from renewable sources.
The latest addition is 60 MW of power from a wind farm in
south Texas close to the city of Corpus Christi. A constant
question is how much of PEC’s generation capacity should be
from renewable sources, especially given the environmental
issues with coal-generated electricity and the rising costs of
hydrocarbon fuels.
Wind and nuclear sources are the current consideration for
the PEC leadership as Texas is increasing its generation by
nuclear power and the state is the national leader in wind
farm–produced electricity.
Consider yourself a member of the board of directors of
PEC. You are an engineer who has been newly elected by the
PEC membership to serve a 3-year term as a director-at-large.
As such, you do not represent a specific district within the
entire service area; all other directors do represent a specific
district. You have many questions about the operations of
PEC, plus you are interested in the economic and societal
benefits of pursuing more renewable source generation
capacity.
Information
Here are some data that you have obtained. The information
is sketchy, as this point, and the numbers are very approximate.
Electricity generation cost estimates are national
in scope, not PEC-specific, and are provided in cents per
kilowatt-hour (¢/kWh).
Generation Cost, ¢/kWh
Fuel Source Likely Range Reasonable Average
Coal 4 to 9 7.4
Natural gas 4 to 10.5 8.6
Wind 4.8 to 9.1 8.2
Solar 4.5 to 15.5 8.8
National average cost of electricity to residential customers:
11¢/kWh
PEC average cost to residential customers: 10.27 ¢/kWh
(from primary sources) and 10.92 ¢/kWh (renewable sources)
Expected life of a generation facility: 20 to 40 years (it is
likely closer to 20 than 40)
Time to construct a facility: 2 to 5 years
Capital cost to build a generation facility: $900 to $1500
per kW
You have also learned that the PEC staff uses the wellrecognized
levelized energy cost (LEC) method to determine
the price of electricity that must be charged to customers to
break even. The formula takes into account the capital cost of
the generation facilities, the cost of capital of borrowed
money, annual maintenance and operation (M&O) costs, and
the expected life of the facility. The LEC formula, expressed
in dollars per kWh for ( t 1, 2, ... , n ), is
tn
P t A t C
—————— t
t1 (1 i)
LEC
t
———————
tn
E
——— t
t1 (1 i) t
where P t capital investments made in year t
A t annual maintenance and operating (M&O) costs
for year t
C t fuel costs for year t
E t amount of electricity generated in year t
n expected life of facility
i discount rate (cost of capital)
Case Study Exercises
1. If you wanted to know more about the new arrangement
with the wind farm in south Texas for the additional
60 MW per year, what types of questions would
you ask of a staff member in your first meeting with
him or her?
2. Much of the current generation capacity of PEC facilities
utilizes coal and natural gas as the primary fuel source.
What about the ethical aspects of the government’s allowance
for these plants to continue polluting the atmosphere
with the emissions that may cause health problems for
citizens and further the effects of global warming? What
types of regulations, if any, should be developed for PEC
(and other generators) to follow in the future?
CASE STUDIES
New and updated case studies at the
end of most chapters present realworld,
in-depth treatments and
exercises in the engineering
profession. Each case includes a
background, relevant information,
and an exercise section.

ACKNOWLEDGMENT OF CONTRIBUTORS
It takes the input and efforts of many individuals to make significant improvements in a textbook.
We wish to give special thanks to the following persons for their contributions to this edition.
Paul Askenasy, Texas Commission on Environmental Quality
Jack Beltran, Bristol-Myers Squibb
Robert Lundquist, Ohio State University
William Peet, Infrastructure Coordination, Government of Niue
Sallie Sheppard, Texas A&M University
We thank the following individuals for their comments, feedback, and review of material to assist
in making this edition a real success.
Ahmed Alim, University of Houston
Alan Atalah, Bowling Green State University
Fola Michael Ayokanmbi, Alabama A&M University
William Brown, West Virginia University at Parkersburg
Hector Carrasco, Colorado State University–Pueblo
Robert Chiang, California State University, Pomona
Ronald Cutwright, Florida State University
John F. Dacquisto, Gonzaga University
Houshang Darabi, University of Illinois at Chicago
Freddie Davis, West Texas A&M University
Edward Lester Dollar, Southern Polytechnic State University
Ted Eschenbach, University of Alaska
Clara Fang, University of Hartford
Abel Fernandez, University of the Pacific
Daniel A. Franchi, California Polytechnic State University, San Luis Obispo
Mark Frascatore, Clarkson University
Benjamin M. Fries, University of Central Florida
Nathan Gartner, University of Massachusetts–Lowell
Johnny R. Graham, University of North Carolina–Charlotte
Liling Huang, Northern Virginia Community College
David Jacobs, University of Hartford
Adam Jannik, Northwestern State University
Peter E. Johnson, Valparaiso University
Justin W. Kile, University of Wisconsin–Platteville
John Kushner, Lawrence Technological University
Clifford D. Madrid, New Mexico State University
Saeed Manafzadeh, University of Illinois at Chicago
Quamrul Mazumder, University of Michigan–Flint
Deb McAvoy, Ohio University
Gene McGinnis, Christian Brothers University
Bruce V. Mutter, Bluefield State College
Hong Sioe Oey, University of Texas at El Paso
Richard Palmer, University of Massachusetts
Michael J. Rider, Ohio Northern University
John Ristroph, University of Louisiana at Lafayette
Saeid L. Sadri, Georgia Institute of Technology
Scott Schultz, Mercer University
Kyo D. Song, Norfolk State University
James Stevens, University of Colorado at Colorado Springs
John A. Stratton, Rochester Institute of Technology
Mathias J. Sutton, Purdue University
Pete Weiss, Valparaiso University

xx
Acknowledgment of Contributors
Greg Wiles, Southern Polytechnic State University
Richard Youchak, University of Pittsburgh at Johnstown
William A. Young, II, Ohio University
If you discover errors that require correction in the next printing of the textbook or in updates of
the online resources, please contact us. We hope you find the contents of this edition helpful in
your academic and professional activities.
Leland Blank
Anthony Tarquin
lelandblank@yahoo.com
atarquin@utep.edu

LEARNING STAGE 1
The Fundamentals
LEARNING STAGE 1
The Fundamentals
CHAPTER 1
Foundations of
Engineering Economy
CHAPTER 2
Factors: How Time
and Interest Affect
Money
CHAPTER 3
Combining Factors
and Spreadsheet
Functions
CHAPTER 4
Nominal and Effective
Interest Rates
The fundamentals of engineering economy are introduced in
these chapters. When you have completed stage 1, you will be
able to understand and work problems that account for the
time value of money, cash flows occurring at different times with
different amounts, and equivalence at different interest rates. The
techniques you master here form the basis of how an engineer in
any discipline can take economic value into account in virtually any
project environment.
The factors commonly used in all engineering economy computations
are introduced and applied here. Combinations of these factors
assist in moving monetary values forward and backward through
time and at different interest rates. Also, after these chapters, you
should be comfortable using many of the spreadsheet functions.
Many of the terms common to economic decision making are
introduced in learning stage 1 and used in later chapters. A checkmark
icon in the margin indicates that a new concept or guideline
is introduced at this point.

CHAPTER 1
Foundations
of Engineering
Economy
LEARNING OUTCOMES
Purpose: Understand and apply fundamental concepts and use the terminology of engineering economics.
SECTION TOPIC LEARNING OUTCOME
1.1 Description and role • Define engineering economics and describe its
role in decision making.
1.2 Engineering economy study
approach
• Understand and identify the steps in an
engineering economy study.
1.3 Ethics and economics • Identify areas in which economic decisions can
present questionable ethics.
1.4 Interest rate • Perform calculations for interest rates and rates
of return.
1.5 Terms and symbols • Identify and use engineering economic
terminology and symbols.
1.6 Cash flows • Understand cash flows and how to graphically
represent them.
1.7 Economic equivalence • Describe and calculate economic equivalence.
1.8 Simple and compound interest • Calculate simple and compound interest
amounts for one or more time periods.
1.9 MARR and opportunity cost • State the meaning and role of Minimum
Attractive Rate of Return (MARR) and
opportunity costs.
1.10 Spreadsheet functions • Identify and use some Excel functions
commonly applied in engineering economics.

T
he need for engineering economy is primarily motivated by the work that engineers
do in performing analyses, synthesizing, and coming to a conclusion as they work on
projects of all sizes. In other words, engineering economy is at the heart of making
decisions . These decisions involve the fundamental elements of cash flows of money, time,
and interest rates. This chapter introduces the basic concepts and terminology necessary for
an engineer to combine these three essential elements in organized, mathematically correct
ways to solve problems that will lead to better decisions.
1.1 Engineering Economics: Description and
Role in Decision Making
Decisions are made routinely to choose one alternative over another by individuals in everyday
life; by engineers on the job; by managers who supervise the activities of others; by corporate
presidents who operate a business; and by government officials who work for the public good.
Most decisions involve money, called capital or capital funds , which is usually limited in
amount. The decision of where and how to invest this limited capital is motivated by a primary
goal of adding value as future, anticipated results of the selected alternative are realized.
Engineers play a vital role in capital investment decisions based upon their ability and experience
to design, analyze, and synthesize. The factors upon which a decision is based are commonly a
combination of economic and noneconomic elements. Engineering economy deals with the
economic factors. By definition,
Engineering economy involves formulating, estimating, and evaluating the expected economic
outcomes of alternatives designed to accomplish a defined purpose. Mathematical techniques
simplify the economic evaluation of alternatives.
Because the formulas and techniques used in engineering economics are applicable to all
types of money matters, they are equally useful in business and government, as well as for
individuals. Therefore, besides applications to projects in your future jobs, what you learn
from this book and in this course may well offer you an economic analysis tool for making
personal decisions such as car purchases, house purchases, major purchases on credit, e.g.,
furniture, appliances, and electronics.
Other terms that mean the same as engineering economy are engineering economic analysis,
capital allocation study, economic analysis, and similar descriptors.
People make decisions; computers, mathematics, concepts, and guidelines assist people in
their decision-making process. Since most decisions affect what will be done, the time frame of
engineering economy is primarily the future . Therefore, the numbers used in engineering economy
are best estimates of what is expected to occur . The estimates and the decision usually
involve four essential elements:
Cash flows
Times of occurrence of cash flows
Interest rates for time value of money
Measure of economic worth for selecting an alternative
Since the estimates of cash flow amounts and timing are about the future, they will be somewhat
different than what is actually observed, due to changing circumstances and unplanned
events. In short, the variation between an amount or time estimated now and that observed
in the future is caused by the stochastic (random) nature of all economic events. Sensitivity
analysis is utilized to determine how a decision might change according to varying estimates,
especially those expected to vary widely. Example 1.1 illustrates the fundamental
nature of variation in estimates and how this variation may be included in the analysis at a
very basic level.
EXAMPLE 1.1
An engineer is performing an analysis of warranty costs for drive train repairs within the first
year of ownership of luxury cars purchased in the United States. He found the average cost (to
the nearest dollar) to be $570 per repair from data taken over a 5-year period.

4 Chapter 1 Foundations of Engineering Economy
Year 2006 2007 2008 2009 2010
Average Cost, $/repair 525 430 619 650 625
What range of repair costs should the engineer use to ensure that the analysis is sensitive to
changing warranty costs?
Solution
At first glance the range should be approximately –25% to 15% of the $570 average cost to
include the low of $430 and high of $650. However, the last 3 years of costs are higher and
more consistent with an average of $631. The observed values are approximately 3% of this
more recent average.
If the analysis is to use the most recent data and trends, a range of, say, 5% of $630 is recommended.
If, however, the analysis is to be more inclusive of historical data and trends, a range
of, say, 20% or 25% of $570 is recommended.
The criterion used to select an alternative in engineering economy for a specific set of estimates
is called a measure of worth . The measures developed and used in this text are
Present worth (PW) Future worth (FW) Annual worth (AW)
Rate of return (ROR) Benefit/cost (B/C) Capitalized cost (CC)
Payback period Economic value added (EVA) Cost Effectiveness
All these measures of worth account for the fact that money makes money over time. This is the
concept of the time value of money.
Time value of money
It is a well-known fact that money makes money. The time value of money explains the change
in the amount of money over time for funds that are owned (invested) or owed (borrowed).
This is the most important concept in engineering economy.
The time value of money is very obvious in the world of economics. If we decide to invest
capital (money) in a project today, we inherently expect to have more money in the future than
we invested. If we borrow money today, in one form or another, we expect to return the original
amount plus some additional amount of money.
Engineering economics is equally well suited for the future and for the analysis of past cash
flows in order to determine if a specific criterion (measure of worth) was attained. For example,
assume you invested $4975 exactly 3 years ago in 53 shares of IBM stock as traded on the New
York Stock Exchange (NYSE) at $93.86 per share. You expect to make 8% per year appreciation,
not considering any dividends that IBM may declare. A quick check of the share value shows it
is currently worth $127.25 per share for a total of $6744.25. This increase in value represents a
rate of return of 10.67% per year. (These type of calculations are explained later.) This past
i nvestment has well exceeded the 8% per year criterion over the last 3 years.
1.2 Performing an Engineering Economy Study
An engineering economy study involves many elements: problem identification, definition of the
objective, cash flow estimation, financial analysis, and decision making. Implementing a structured
procedure is the best approach to select the best solution to the problem.
The steps in an engineering economy study are as follows:
1. Identify and understand the problem; identify the objective of the project.
2. Collect relevant, available data and define viable solution alternatives.
3. Make realistic cash flow estimates.
4. Identify an economic measure of worth criterion for decision making.

1.2 Performing an Engineering Economy Study 5
5. Evaluate each alternative; consider noneconomic factors; use sensitivity analysis as needed.
6. Select the best alternative.
7. Implement the solution and monitor the results.
Technically, the last step is not part of the economy study, but it is, of course, a step needed to
meet the project objective. There may be occasions when the best economic alternative
requires more capital funds than are available, or significant noneconomic factors preclude the
most economic alternative from being chosen. Accordingly, steps 5 and 6 may result in selection
of an alternative different from the economically best one. Also, sometimes more than one project
may be selected and implemented. This occurs when projects are independent of one another.
In this case, steps 5 through 7 vary from those above. Figure 1–1 illustrates the steps above for
one alternative. Descriptions of several of the elements in the steps are important to understand.
Problem Description and Objective Statement A succinct statement of the problem and
primary objective(s) is very important to the formation of an alternative solution. As an illustration,
assume the problem is that a coal-fueled power plant must be shut down by 2015 due to the
production of excessive sulfur dioxide. The objectives may be to generate the forecasted electricity
Step in
study
1
Problem
description
Objective
statement
2
Available data
Alternatives for
solution
One or more approaches
to meet objective
3
Cash flows and
other estimates
Expected life
Revenues
Costs
Taxes
Project financing
4
Measure of worth
criterion
PW, ROR, B/C, etc.
5
6
Engineering
economic analysis
Best alternative
selection
Consider:
• Noneconomic factors
• Sensitivity analysis
• Risk analysis
7
Implementation
and monitoring
Time
passes
1
New problem
description
New engineering
economy study
begins
Figure 1–1
Steps in an engineering economy study.

6 Chapter 1 Foundations of Engineering Economy
needed for 2015 and beyond, plus to not exceed all the projected emission allowances in these
future years.
Alternatives These are stand-alone descriptions of viable solutions to problems that can meet
the objectives. Words, pictures, graphs, equipment and service descriptions, simulations, etc.
define each alternative. The best estimates for parameters are also part of the alternative. Some
parameters include equipment first cost, expected life, salvage value (estimated trade-in, resale,
or market value), and annual operating cost (AOC), which can also be termed maintenance and
operating (M&O) cost, and subcontract cost for specific services. If changes in income (revenue)
may occur, this parameter must be estimated.
Detailing all viable alternatives at this stage is crucial. For example, if two alternatives are
described and analyzed, one will likely be selected and implementation initiated. If a third, more
attractive method that was available is later recognized, a wrong decision was made.
Cash Flows All cash flows are estimated for each alternative. Since these are future expenditures
and revenues, the results of step 3 usually prove to be inaccurate when an alternative is
actually in place and operating. When cash flow estimates for specific parameters are expected to
vary significantly from a point estimate made now, risk and sensitivity analyses (step 5) are
needed to improve the chances of selecting the best alternative. Sizable variation is usually expected
in estimates of revenues, AOC, salvage values, and subcontractor costs. Estimation of
costs is discussed in Chapter 15, and the elements of variation (risk) and sensitivity analysis are
included throughout the text.
Engineering Economy Analysis The techniques and computations that you will learn and
use throughout this text utilize the cash flow estimates, time value of money, and a selected
measure of worth. The result of the analysis will be one or more numerical values; this can be
in one of several terms, such as money, an interest rate, number of years, or a probability. In
the end, a selected measure of worth mentioned in the previous section will be used to select
the best alternative.
Before an economic analysis technique is applied to the cash flows, some decisions about
what to include in the analysis must be made. Two important possibilities are taxes and
inflation. Federal, state or provincial, county, and city taxes will impact the costs of every
alternative. An after-tax analysis includes some additional estimates and methods compared to
a before-tax a nalysis. If taxes and inflation are expected to impact all alternatives equally, they
may be disregarded in the analysis. However, if the size of these projected costs is important,
taxes and inflation should be considered. Also, if the impact of inflation over time is important
to the decision, an additional set of computations must be added to the analysis; Chapter 14
covers the details.
Selection of the Best Alternative The measure of worth is a primary basis for selecting
the best economic alternative. For example, if alternative A has a rate of return (ROR) of
15.2% per year and alternative B will result in an ROR of 16.9% per year, B is better economically.
However, there can always be noneconomic or intangible factors that must be
considered and that may alter the decision. There are many possible noneconomic factors;
some typical ones are
• Market pressures, such as need for an increased international presence
• Availability of certain resources, e.g., skilled labor force, water, power, tax incentives
• Government laws that dictate safety, environmental, legal, or other aspects
• Corporate management’s or the board of director’s interest in a particular alternative
• Goodwill offered by an alternative toward a group: employees, union, county, etc.
As indicated in Figure 1–1 , once all the economic, noneconomic, and risk factors have been
evaluated, a final decision of the “best” alternative is made.
At times, only one viable alternative is identified. In this case, the do-nothing (DN) alternative
may be chosen provided the measure of worth and other factors result in the alternative being
a poor choice. The do-nothing alternative maintains the status quo.

1.3 Professional Ethics and Economic Decisions 7
Whether we are aware of it or not, we use criteria every day to choose between alternatives.
For example, when you drive to campus, you decide to take the “best” route. But how did you
define best? Was the best route the safest, shortest, fastest, cheapest, most scenic, or what? Obviously,
depending upon which criterion or combination of criteria is used to identify the best, a
different route might be selected each time. In economic analysis, financial units (dollars or
other currency) are generally used as the tangible basis for evaluation. Thus, when there are
several ways of accomplishing a stated objective, the alternative with the lowest overall cost or
highest overall net income is selected.
1.3 Professional Ethics and Economic Decisions
Many of the fundamentals of engineering ethics are intertwined with the roles of money and
economics-based decisions in the making of professionally ethical judgments. Some of these
integral connections are discussed here, plus sections in later chapters discuss additional aspects
of ethics and economics. For example, Chapter 9, Benefit/Cost Analysis and Public Sector Economics,
includes material on the ethics of public project contracts and public policy. Although it
is very limited in scope and space, it is anticipated that this coverage of the important role of
economics in engineering ethics will prompt further interest on the part of students and instructors
of engineering economy.
The terms morals and ethics are commonly used interchangeably, yet they have slightly
different interpretations. Morals usually relate to the underlying tenets that form the character
and conduct of a person in judging right and wrong. Ethical practices can be evaluated by
using a code of morals or code of ethics that forms the standards to guide decisions and
actions of individuals and organizations in a profession, for example, electrical, chemical,
mechanical, industrial, or civil engineering. There are several different levels and types of
morals and ethics.
Universal or common morals These are fundamental moral beliefs held by virtually all people.
Most people agree that to steal, murder, lie, or physically harm someone is wrong.
It is possible for actions and intentions to come into conflict concerning a common moral.
Consider the World Trade Center buildings in New York City. After their collapse on September 11,
2001, it was apparent that the design was not sufficient to withstand the heat generated by the
firestorm caused by the impact of an aircraft. The structural engineers who worked on the design
surely did not have the intent to harm or kill occupants in the buildings. However, their design
actions did not foresee this outcome as a measurable possibility. Did they violate the common
moral belief of not doing harm to others or murdering?
Individual or personal morals These are the moral beliefs that a person has and maintains
over time. These usually parallel the common morals in that stealing, lying, murdering, etc. are
immoral acts.
It is quite possible that an individual strongly supports the common morals and has excellent
personal morals, but these may conflict from time to time when decisions must be made. Consider
the engineering student who genuinely believes that cheating is wrong. If he or she does not
know how to work some test problems, but must make a certain minimum grade on the final
exam to graduate, the decision to cheat or not on the final exam is an exercise in following or
violating a personal moral.
Professional or engineering ethics Professionals in a specific discipline are guided in their
decision making and performance of work activities by a formal standard or code. The code
states the commonly accepted standards of honesty and integrity that each individual is expected
to demonstrate in her or his practice. There are codes of ethics for medical doctors, attorneys,
and, of course, engineers.
Although each engineering profession has its own code of ethics, the Code of Ethics for
Engineers published by the National Society of Professional Engineers (NSPE) is very commonly
used and quoted. This code, reprinted in its entirety in Appendix C, includes numerous
sections that have direct or indirect economic and financial impact upon the designs, actions,

8 Chapter 1 Foundations of Engineering Economy
and decisions that engineers make in their professional dealings. Here are three examples from
the Code:
“Engineers, in the fulfillment of their duties, shall hold paramount the safety, health, and welfare
of the public .” (section I.1)
“Engineers shall not accept fi nancial or other considerations , including free engineering designs,
from material or equipment suppliers for specifying their product.” (section III.5.a)
“Engineers using designs supplied by a client recognize that the designs remain the property
of the client and may not be duplicated by the engineer for others without express permission.”
(section III.9.b)
As with common and personal morals, conflicts can easily rise in the mind of an engineer
between his or her own ethics and that of the employing corporation. Consider a manufacturing
engineer who has recently come to firmly disagree morally with war and its negative effects on
human beings. Suppose the engineer has worked for years in a military defense contractor’s
facility and does the detailed cost estimations and economic evaluations of producing fighter
jets for the Air Force. The Code of Ethics for Engineers is silent on the ethics of producing and
using war materiel. Although the employer and the engineer are not violating any ethics code,
the engineer, as an individual, is stressed in this position. Like many people during a declining
national economy, retention of this job is of paramount importance to the family and the engineer.
Conflicts such as this can place individuals in real dilemmas with no or mostly unsatisfactory
alternatives.
At first thought, it may not be apparent how activities related to engineering economics may
present an ethical challenge to an individual, a company, or a public servant in government service.
Many money-related situations, such as those that follow, can have ethical dimensions.
In the design stage:
• Safety factors are compromised to ensure that a price bid comes in as low as possible.
• Family or personal connections with individuals in a company offer unfair or insider information
that allows costs to be cut in strategic areas of a project.
• A potential vendor offers specifications for company-specific equipment, and the design engineer
does not have sufficient time to determine if this equipment will meet the needs of the
project being designed and costed.
While the system is operating:
• Delayed or below-standard maintenance can be performed to save money when cost overruns
exist in other segments of a project.
• Opportunities to purchase cheaper repair parts can save money for a subcontractor working on
a fixed-price contract.
• Safety margins are compromised because of cost, personal inconvenience to workers, tight
time schedules, etc.
A good example of the last item—safety is compromised while operating the system—is the
situation that arose in 1984 in Bhopal, India (Martin and Schinzinger 2005, pp. 245–8). A Union
Carbide plant manufacturing the highly toxic pesticide chemical methyl isocyanate (MIC) experienced
a large gas leak from high-pressure tanks. Some 500,000 persons were exposed to inhalation
of this deadly gas that burns moist parts of the body. There were 2500 to 3000 deaths within
days, and over the following 10-year period, some 12,000 death claims and 870,000 personal
injury claims were recorded. Although Union Carbide owned the facility, the Indian government
had only Indian workers in the plant. Safety practices clearly eroded due to cost-cutting measures,
insufficient repair parts, and reduction in personnel to save salary money. However, one of
the surprising practices that caused unnecessary harm to workers was the fact that masks, gloves,
and other protective gear were not worn by workers in close proximity to the tanks containing
MIC. Why? Unlike in plants in the United States and other countries, there was no air conditioning
in the Indian plant, resulting in high ambient temperatures in the facility.
Many ethical questions arise when corporations operate in international settings where the
corporate rules, worker incentives, cultural practices, and costs in the home country differ from
those in the host country. Often these ethical dilemmas are fundamentally based in the economics
that provide cheaper labor, reduced raw material costs, less government oversight, and a host of

1.3 Professional Ethics and Economic Decisions 9
other cost-reducing factors. When an engineering economy study is performed, it is important for
the engineer performing the study to consider all ethically related matters to ensure that the cost
and revenue estimates reflect what is likely to happen once the project or system is operating.
It is important to understand that the translation from universal morals to personal morals and
professional ethics does vary from one culture and country to another. As an example, consider the
common belief (universal moral) that the awarding of contracts and financial arrangements for services
to be performed (for government or business) should be accomplished in a fair and transparent
fashion. In some societies and cultures, corruption in the process of contract making is common and
often “overlooked” by the local authorities, who may also be involved in the affairs. Are these immoral
or unethical practices? Most would say, “Yes, this should not be allowed. Find and punish the
individuals involved.” Yet, such practices do continue, thus indicating the differences in interpretation
of common morals as they are translated into the ethics of individuals and professionals.
EXAMPLE 1.2
Jamie is an engineer employed by Burris, a United States–based company that develops subway
and surface transportation systems for medium-sized municipalities in the United States
and Canada. He has been a registered professional engineer (PE) for the last 15 years. Last
year, Carol, an engineer friend from university days who works as an individual consultant,
asked Jamie to help her with some cost estimates on a metro train job. Carol offered to pay for
his time and talent, but Jamie saw no reason to take money for helping with data commonly
used by him in performing his job at Burris. The estimates took one weekend to complete, and
once Jamie delivered them to Carol, he did not hear from her again; nor did he learn the identity
of the company for which Carol was preparing the estimates.
Yesterday, Jamie was called into his supervisor’s office and told that Burris had not received
the contract award in Sharpstown, where a metro system is to be installed. The project estimates
were prepared by Jamie and others at Burris over the past several months. This job was
greatly needed by Burris, as the country and most municipalities were in a real economic
slump, so much so that Burris was considering furloughing several engineers if the Sharpstown
bid was not accepted. Jamie was told he was to be laid off immediately, not because the bid was
rejected, but because he had been secretly working without management approval for a prime
consultant of Burris’ main competitor. Jamie was astounded and angry. He knew he had done
nothing to warrant firing, but the evidence was clearly there. The numbers used by the competitor
to win the Sharpstown award were the same numbers that Jamie had prepared for Burris
on this bid, and they closely matched the values that he gave Carol when he helped her.
Jamie was told he was fortunate, because Burris’ president had decided to not legally charge
Jamie with unethical behavior and to not request that his PE license be rescinded. As a result,
Jamie was escorted out of his office and the building within one hour and told to not ask anyone
at Burris for a reference letter if he attempted to get another engineering job.
Discuss the ethical dimensions of this situation for Jamie, Carol, and Burris’ management.
Refer to the NSPE Code of Ethics for Engineers (Appendix C) for specific points of concern.
Solution
There are several obvious errors and omissions present in the actions of Jamie, Carol, and
B urris’ management in this situation. Some of these mistakes, oversights, and possible code
violations are summarized here.
Jamie
• Did not learn identity of company Carol was working for and whether the company was to
be a bidder on the Sharpstown project
• Helped a friend with confidential data, probably innocently, without the knowledge or approval
of his employer
• Assisted a competitor, probably unknowingly, without the knowledge or approval of his
employer
• Likely violated, at least, Code of Ethics for Engineers section II.1.c, which reads, “Engineers
shall not reveal facts, data, or information without the prior consent of the client or
employer except as authorized or required by law or this Code.”

10 Chapter 1 Foundations of Engineering Economy
Carol
• Did not share the intended use of Jamie’s work
• Did not seek information from Jamie concerning his employer’s intention to bid on the
same project as her client
• Misled Jamie in that she did not seek approval from Jamie to use and quote his information
and assistance
• Did not inform her client that portions of her work originated from a source employed by a
possible bid competitor
• Likely violated, at least, Code of Ethics for Engineers section III.9.a, which reads, “Engineers
shall, whenever possible, name the person or persons who may be individually responsible
for designs, inventions, writings, or other accomplishments.”
Burris’ management
• Acted too fast in dismissing Jamie; they should have listened to Jamie and conducted an
investigation
• Did not put him on administrative leave during a review
• Possibly did not take Jamie’s previous good work record into account
These are not all ethical considerations; some are just plain good business practices for Jamie,
Carol, and Burris.
1.4 Interest Rate and Rate of Return
Interest is the manifestation of the time value of money. Computationally, interest is the difference
between an ending amount of money and the beginning amount. If the difference is zero or negative,
there is no interest. There are always two perspectives to an amount of interest—interest paid
and interest earned. These are illustrated in Figure 1–2 . Interest is paid when a person or organization
borrowed money (obtained a loan) and repays a larger amount over time. Interest is earned
when a person or organization saved, invested, or lent money and obtains a return of a larger
amount over time. The numerical values and formulas used are the same for both perspectives, but
the interpretations are different.
Interest paid on borrowed funds (a loan) is determined using the original amount, also called
the principal,
Interest amount owed now principal [1.1]
When interest paid over a specifi c time unit is expressed as a percentage of the principal, the result
is called the interest rate.
interest accrued per time unit
Interest rate (%) ————————————— 100% [1.2]
principal
The time unit of the rate is called the interest period. By far the most common interest period
used to state an interest rate is 1 year. Shorter time periods can be used, such as 1% per month.
Thus, the interest period of the interest rate should always be included. If only the rate is stated,
for example, 8.5%, a 1-year interest period is assumed.
Loan
Loan
Bank
Repayment
interest
Borrower
Investor
(a)
Figure 1–2
(a) Interest paid over time to lender. (b) Interest earned over time by investor.
Repayment
interest
(b)
Corporation

1.4 Interest Rate and Rate of Return 11
EXAMPLE 1.3
An employee at LaserKinetics.com borrows $10,000 on May 1 and must repay a total of
$10,700 exactly 1 year later. Determine the interest amount and the interest rate paid.
Solution
The perspective here is that of the borrower since $10,700 repays a loan. Apply Equation [1.1]
to determine the interest paid.
Interest paid $10,700 10,000 $700
Equation [1.2] determines the interest rate paid for 1 year.
$700
Percent interest rate ———— 100% 7% per year
$10,000
EXAMPLE 1.4
Stereophonics, Inc., plans to borrow $20,000 from a bank for 1 year at 9% interest for new
recording equipment. ( a ) Compute the interest and the total amount due after 1 year. ( b ) Construct
a column graph that shows the original loan amount and total amount due after 1 year
used to compute the loan interest rate of 9% per year.
Solution
(a) Compute the total interest accrued by solving Equation [1.2] for interest accrued.
Interest $20,000(0.09) $1800
The total amount due is the sum of principal and interest.
Total due $20,000 1800 $21,800
(b) Figure 1–3 shows the values used in Equation [1.2]: $1800 interest, $20,000 original loan
principal, 1-year interest period.
$
$21,800
$20,000
Original
loan
amount
Interest = $1800
Interest rate
$1800
$20,000 100%
= 9% per year
Now
1 year
later
Interest
period is
1 year
Figure 1–3
Values used to compute an interest rate of 9% per year. Example 1.4.
Comment
Note that in part ( a ), the total amount due may also be computed as
Total due principal(1 interest rate) $20,000(1.09) $21,800
Later we will use this method to determine future amounts for times longer than one interest
period.

12 Chapter 1 Foundations of Engineering Economy
From the perspective of a saver, a lender, or an investor, interest earned (Figure 1–2 b ) is the
final amount minus the initial amount, or principal.
Interest earned total amount now principal [1.3]
Interest earned over a specific period of time is expressed as a percentage of the original amount
and is called rate of return (ROR).
interest accrued per time unit
Rate of return (%) ————————————— 100% [1.4]
principal
The time unit for rate of return is called the interest period, just as for the borrower’s perspective.
Again, the most common period is 1 year.
The term return on investment (ROI) is used equivalently with ROR in different industries and
settings, especially where large capital funds are committed to engineering-oriented programs.
The numerical values in Equations [1.2] and [1.4] are the same, but the term interest rate paid
is more appropriate for the borrower’s perspective, while the rate of return earned is better for
the investor’s perspective.
EXAMPLE 1.5
(a) Calculate the amount deposited 1 year ago to have $1000 now at an interest rate of 5%
per year.
(b) Calculate the amount of interest earned during this time period.
Solution
(a) The total amount accrued ($1000) is the sum of the original deposit and the earned interest.
If X is the original deposit,
Total accrued deposit deposit(interest rate)
$1000 X X (0.05) X (1 0.05) 1.05 X
The original deposit is
X ——— 1000
1.05 $952.38
(b) Apply Equation [1.3] to determine the interest earned.
Interest $1000 952.38 $47.62
In Examples 1.3 to 1.5 the interest period was 1 year, and the interest amount was calculated
at the end of one period. When more than one interest period is involved, e.g., the amount of interest
after 3 years, it is necessary to state whether the interest is accrued on a simple or compound
basis from one period to the next. This topic is covered later in this chapter.
Since inflation can significantly increase an interest rate, some comments about the fundamentals
of inflation are warranted at this early stage. By definition, inflation represents a decrease
in the value of a given currency. That is, $10 now will not purchase the same amount of gasoline
for your car (or most other things) as $10 did 10 years ago. The changing value of the currency
affects market interest rates.
Inflation
In simple terms, interest rates reflect two things: a so-called real rate of return plus the expected
inflation rate. The real rate of return allows the investor to purchase more than he or she could
have purchased before the investment, while inflation raises the real rate to the market rate that
we use on a daily basis.
The safest investments (such as government bonds) typically have a 3% to 4% real rate of
return built into their overall interest rates. Thus, a market interest rate of, say, 8% per year on a
bond means that investors expect the inflation rate to be in the range of 4% to 5% per year.
Clearly, inflation causes interest rates to rise.
From the borrower’s perspective, the rate of inflation is another interest rate tacked on to the
real interest rate . And from the vantage point of the saver or investor in a fixed-interest account,

1.5 Terminology and Symbols 13
inflation reduces the real rate of return on the investment. Inflation means that cost and revenue
cash flow estimates increase over time. This increase is due to the changing value of money that
is forced upon a country’s currency by inflation, thus making a unit of currency (such as the dollar)
worth less relative to its value at a previous time. We see the effect of inflation in that money
purchases less now than it did at a previous time. Inflation contributes to
• A reduction in purchasing power of the currency
• An increase in the CPI (consumer price index)
• An increase in the cost of equipment and its maintenance
• An increase in the cost of salaried professionals and hourly employees
• A reduction in the real rate of return on personal savings and certain corporate investments
In other words, inflation can materially contribute to changes in corporate and personal economic
analysis.
Commonly, engineering economy studies assume that inflation affects all estimated values
equally. Accordingly, an interest rate or rate of return, such as 8% per year, is applied throughout
the analysis without accounting for an additional inflation rate. However, if inflation were explicitly
taken into account, and it was reducing the value of money at, say, an average of 4% per year,
then it would be necessary to perform the economic analysis using an inflated interest rate. (The
rate is 12.32% per year using the relations derived in Chapter 14.)
1.5 Terminology and Symbols
The equations and procedures of engineering economy utilize the following terms and symbols.
Sample units are indicated.
P value or amount of money at a time designated as the present or time 0. Also P is
referred to as present worth (PW), present value (PV), net present value (NPV), discounted
cash flow (DCF), and capitalized cost (CC); monetary units, such as dollars
F value or amount of money at some future time. Also F is called future worth (FW)
and future value (FV); dollars
A series of consecutive, equal, end-of-period amounts of money. Also A is called the
annual worth (AW) and equivalent uniform annual worth (EUAW); dollars per
year, euros per month
n number of interest periods; years, months, days
i interest rate per time period; percent per year, percent per month
t time, stated in periods; years, months, days
The symbols P and F represent one-time occurrences: A occurs with the same value in each interest
period for a specified number of periods. It should be clear that a present value P represents a
single sum of money at some time prior to a future value F or prior to the first occurrence of an
equivalent series amount A .
It is important to note that the symbol A always represents a uniform amount (i.e., the same
amount each period) that extends through consecutive interest periods. Both conditions must
exist before the series can be represented by A .
The interest rate i is expressed in percent per interest period, for example, 12% per year. Unless
stated otherwise, assume that the rate applies throughout the entire n years or interest periods.
The decimal equivalent for i is always used in formulas and equations in engineering economy
computations.
All engineering economy problems involve the element of time expressed as n and interest
rate i . In general, every problem will involve at least four of the symbols P , F , A , n , and i , with at
least three of them estimated or known.
Additional symbols used in engineering economy are defined in Appendix E.
EXAMPLE 1.6
Today, Julie borrowed $5000 to purchase furniture for her new house. She can repay the loan
in either of the two ways described below. Determine the engineering economy symbols and
their value for each option.

14 Chapter 1 Foundations of Engineering Economy
(a) Five equal annual installments with interest based on 5% per year.
(b) One payment 3 years from now with interest based on 7% per year.
Solution
(a) The repayment schedule requires an equivalent annual amount A , which is unknown.
P $5000 i 5% per year n 5 years A ?
(b) Repayment requires a single future amount F, which is unknown.
P $5000 i 7% per year n 3 years F ?
EXAMPLE 1.7
You plan to make a lump-sum deposit of $5000 now into an investment account that pays 6%
per year, and you plan to withdraw an equal end-of-year amount of $1000 for 5 years, starting
next year. At the end of the sixth year, you plan to close your account by withdrawing the remaining
money. Define the engineering economy symbols involved.
Solution
All five symbols are present, but the future value in year 6 is the unknown.
P $5000
A $1000 per year for 5 years
F ? at end of year 6
i 6% per year
n 5 years for the A series and 6 for the F value
EXAMPLE 1.8
Last year Jane’s grandmother offered to put enough money into a savings account to generate
$5000 in interest this year to help pay Jane’s expenses at college. ( a ) Identify the symbols, and
( b ) calculate the amount that had to be deposited exactly 1 year ago to earn $5000 in interest
now, if the rate of return is 6% per year.
Solution
(a) Symbols P (last year is 1) and F (this year) are needed.
P ?
i 6% per year
n 1 year
F P interest ? $5000
(b) Let F total amount now and P original amount. We know that F – P $5000 is
accrued interest. Now we can determine P . Refer to Equations [1.1] through [1.4].
F P Pi
The $5000 interest can be expressed as
Interest F – P ( P Pi ) – P
Pi
$5000 P (0.06)
P $5000 ———
0.06 $83,333.33

1.6 Cash Flows: Estimation and Diagramming 15
1.6 Cash Flows: Estimation and Diagramming
As mentioned in earlier sections, cash flows are the amounts of money estimated for future projects
or observed for project events that have taken place. All cash flows occur during specific time periods,
such as 1 month, every 6 months, or 1 year. Annual is the most common time period. For
example, a payment of $10,000 once every year in December for 5 years is a series of 5 outgoing
cash flows. And an estimated receipt of $500 every month for 2 years is a series of 24 incoming cash
flows. Engineering economy bases its computations on the timing, size, and direction of cash flows.
Cash inflows are the receipts, revenues, incomes, and savings generated by project and business
activity. A plus sign indicates a cash inflow.
Cash outflows are costs, disbursements, expenses, and taxes caused by projects and business
activity. A negative or minus sign indicates a cash outflow. When a project involves only costs,
the minus sign may be omitted for some techniques, such as benefit/cost analysis.
Cash flow
Of all the steps in Figure 1–1 that outline the engineering economy study, estimating cash flows
(step 3) is the most difficult, primarily because it is an attempt to predict the future. Some examples
of cash flow estimates are shown here. As you scan these, consider how the cash inflow
or outflow may be estimated most accurately.
Cash Inflow Estimates
Income: $150,000 per year from sales of solar-powered watches
Savings: $24,500 tax savings from capital loss on equipment salvage
Receipt: $750,000 received on large business loan plus accrued interest
Savings: $150,000 per year saved by installing more efficient air conditioning
Revenue: $50,000 to $75,000 per month in sales for extended battery life iPhones
Cash Outflow Estimates
Operating costs: $230,000 per year annual operating costs for software services
First cost: $800,000 next year to purchase replacement earthmoving equipment
Expense: $20,000 per year for loan interest payment to bank
Initial cost: $1 to $1.2 million in capital expenditures for a water recycling unit
All of these are point estimates , that is, single-value estimates for cash flow elements of an
alternative, except for the last revenue and cost estimates listed above. They provide a range estimate,
because the persons estimating the revenue and cost do not have enough knowledge or experience
with the systems to be more accurate. For the initial chapters, we will utilize point estimates. The use
of risk and sensitivity analysis for range estimates is covered in the later chapters of this book.
Once all cash inflows and outflows are estimated (or determined for a completed project), the
net cash flow for each time period is calculated.
Net cash flow cash inflows cash outflows [1.5]
NCF R D [1.6]
where NCF is net cash flow, R is receipts, and D is disbursements.
At the beginning of this section, the timing, size, and direction of cash fl ows were mentioned
as important. Because cash flows may take place at any time during an interest period, as a matter
of convention, all cash flows are assumed to occur at the end of an interest period.
The end-of-period convention means that all cash inflows and all cash outflows are assumed to
take place at the end of the interest period in which they actually occur. When several inflows
and outflows occur within the same period, the net cash flow is assumed to occur at the end of
the period.
End-of-period convention

16 Chapter 1 Foundations of Engineering Economy
Figure 1–4
A typical cash flow time
scale for 5 years.
Year 1
0 1
2
3 4
Time scale
Year 5
5
Figure 1–5
Example of positive and
negative cash flows.
+
i = 4% per year
F =?
Cash flow
1
2
3
4
5
Year
–
In assuming end-of-period cash flows, it is important to understand that future (F) and uniform
annual (A) amounts are located at the end of the interest period, which is not necessarily
December 31. If in Example 1.7 the lump-sum deposit took place on July 1, 2011, the withdrawals
will take place on July 1 of each succeeding year for 6 years. Remember, end of the period
means end of interest period, not end of calendar year.
The cash flow diagram is a very important tool in an economic analysis, especially when the
cash flow series is complex. It is a graphical representation of cash flows drawn on the y axis with
a time scale on the x axis. The diagram includes what is known, what is estimated, and what is
needed. That is, once the cash flow diagram is complete, another person should be able to work
the problem by looking at the diagram.
Cash flow diagram time t 0 is the present, and t 1 is the end of time period 1. We assume
that the periods are in years for now. The time scale of Figure 1–4 is set up for 5 years. Since the
end-of-year convention places cash flows at the ends of years, the “1” marks the end of year 1.
While it is not necessary to use an exact scale on the cash flow diagram, you will probably
avoid errors if you make a neat diagram to approximate scale for both time and relative cash flow
magnitudes.
The direction of the arrows on the diagram is important to differentiate income from outgo. A
vertical arrow pointing up indicates a positive cash flow. Conversely, a down-pointing arrow indicates
a negative cash flow. We will use a bold, colored arrow to indicate what is unknown
and to be determined. For example, if a future value F is to be determined in year 5, a wide,
colored arrow with F ? is shown in year 5. The interest rate is also indicated on the diagram.
Figure 1–5 illustrates a cash inflow at the end of year 1, equal cash outflows at the end of years 2
and 3, an interest rate of 4% per year, and the unknown future value F after 5 years. The arrow
for the unknown value is generally drawn in the opposite direction from the other cash flows;
however, the engineering economy computations will determine the actual sign on the F value.
Before the diagramming of cash flows, a perspective or vantage point must be determined so
that or – signs can be assigned and the economic analysis performed correctly. Assume you
borrow $8500 from a bank today to purchase an $8000 used car for cash next week, and you plan
to spend the remaining $500 on a new paint job for the car two weeks from now. There are several
perspectives possible when developing the cash flow diagram—those of the borrower (that’s
you), the banker, the car dealer, or the paint shop owner. The cash flow signs and amounts for
these perspectives are as follows.
Perspective Activity Cash flow with Sign, $ Time, week
You Borrow 8500 0
Buy car −8000 1
Paint job −500 2
Banker Lender −8500 0
Car dealer Car sale 8000 1
Painter Paint job 500 2

1.6 Cash Flows: Estimation and Diagramming 17
$8500
1
2
0
$500
Week
$8000
Figure 1–6
Cash flows from perspective of borrower for loan and purchases.
One, and only one, of the perspectives is selected to develop the diagram. For your perspective,
all three cash flows are involved and the diagram appears as shown in Figure 1–6 with a time scale
of weeks. Applying the end-of-period convention, you have a receipt of $8500 now (time 0) and
cash outflows of $8000 at the end of week 1, followed by $500 at the end of week 2.
EXAMPLE 1.9
Each year Exxon-Mobil expends large amounts of funds for mechanical safety features
throughout its worldwide operations. Carla Ramos, a lead engineer for Mexico and Central
American operations, plans expenditures of $1 million now and each of the next 4 years just
for the improvement of field-based pressure-release valves. Construct the cash flow diagram to
find the equivalent value of these expenditures at the end of year 4, using a cost of capital estimate
for safety-related funds of 12% per year.
Solution
Figure 1–7 indicates the uniform and negative cash flow series (expenditures) for five periods,
and the unknown F value (positive cash flow equivalent) at exactly the same time as the fifth
expenditure. Since the expenditures start immediately, the first $1 million is shown at time 0,
not time 1. Therefore, the last negative cash flow occurs at the end of the fourth year, when F
also occurs. To make this diagram have a full 5 years on the time scale, the addition of the
year 1 completes the diagram. This addition demonstrates that year 0 is the end-of-period
point for the year 1.
i = 12%
F =?
1
0 1 2
3
4
Year
A = $1,000,000
Figure 1–7
Cash flow diagram, Example 1.9.
EXAMPLE 1.10
An electrical engineer wants to deposit an amount P now such that she can withdraw an equal
annual amount of A 1 $2000 per year for the first 5 years, starting 1 year after the deposit, and
a different annual withdrawal of A 2 $3000 per year for the following 3 years. How would the
cash flow diagram appear if i 8.5% per year?

18 Chapter 1 Foundations of Engineering Economy
Solution
The cash flows are shown in Figure 1–8. The negative cash outflow P occurs now. The withdrawals
(positive cash inflow) for the A 1 series occur at the end of years 1 through 5, and A 2
occurs in years 6 through 8.
A 1 = $2000
A 2 = $3000
0 1 2 3 4 5 6 7
i = 8.5%
8
Year
P =?
Figure 1–8
Cash flow diagram with two different A series, Example 1.10.
EXAMPLE 1.11
A rental company spent $2500 on a new air compressor 7 years ago. The annual rental income
from the compressor has been $750. The $100 spent on maintenance the first year has increased
each year by $25. The company plans to sell the compressor at the end of next year for
$150. Construct the cash flow diagram from the company’s perspective and indicate where the
present worth now is located.
Solution
Let now be time t 0. The incomes and costs for years 7 through 1 (next year) are tabulated
below with net cash flow computed using Equation [1.5]. The net cash flows (one negative,
eight positive) are diagrammed in Figure 1–9 . Present worth P is located at year 0.
End of Year Income Cost Net Cash Flow
−7 $ 0 $2500 $−2500
−6 750 100 650
−5 750 125 625
−4 750 150 600
−3 750 175 575
−2 750 200 550
−1 750 225 525
0 750 250 500
1 750 150 275 625
$650
$625
$600
$575
$550
$525
P =?
$500
$625
–7 –6 –5 –4 –3 –2 –1 0
1
Year
$2500
Figure 1–9
Cash flow diagram, Example 1.11.

1.7 Economic Equivalence 19
1.7 Economic Equivalence
Economic equivalence is a fundamental concept upon which engineering economy computations
are based. Before we delve into the economic aspects, think of the many types of equivalency we
may utilize daily by transferring from one scale to another. Some example transfers between
scales are as follows:
Length:
12 inches 1 foot 3 feet 1 yard 39.370 inches 1 meter
100 centimeters 1 meter 1000 meters 1 kilometer 1 kilometer 0.621 mile
Pressure:
1 atmosphere 1 newton/meter 2 10 3 pascal 1 kilopascal
Often equivalency involves two or more scales. Consider the equivalency of a speed of 110 kilometers
per hour (kph) into miles per minute using conversions between distance and time scales
with three-decimal accuracy.
Speed:
1 mile 1.609 kilometers 1 hour 60 minutes
110 kph 68.365 miles per hour (mph) 68.365 mph 1.139 miles per minute
Four scales—time in minutes, time in hours, length in miles, and length in kilometers—are
combined to develop these equivalent statements on speed. Note that throughout these statements,
the fundamental relations of 1 mile 1.609 kilometers and 1 hour 60 minutes are
applied. If a fundamental relation changes, the entire equivalency is in error.
Now we consider economic equivalency.
Economic equivalence is a combination of interest rate and time value of money to determine
the different amounts of money at different points in time that are equal in economic
value.
Economic equivalence
As an illustration, if the interest rate is 6% per year, $100 today (present time) is equivalent to
$106 one year from today.
Amount accrued 100 100(0.06) 100(1 0.06) $106
If someone offered you a gift of $100 today or $106 one year from today, it would make no difference
which offer you accepted from an economic perspective. In either case you have $106
one year from today. However, the two sums of money are equivalent to each other only when the
interest rate is 6% per year. At a higher or lower interest rate, $100 today is not equivalent to $106
one year from today.
In addition to future equivalence, we can apply the same logic to determine equivalence for
previous years. A total of $100 now is equivalent to $1001.06 $94.34 one year ago at an
interest rate of 6% per year. From these illustrations, we can state the following: $94.34 last
year, $100 now, and $106 one year from now are equivalent at an interest rate of 6% per year.
The fact that these sums are equivalent can be verified by computing the two interest rates for
1-year interest periods.
$6
——— 100% 6% per year
$100
and
$5.66
——— 100% 6% per year
$94.34
The cash flow diagram in Figure 1–10 indicates the amount of interest needed each year to make
these three different amounts equivalent at 6% per year.

20 Chapter 1 Foundations of Engineering Economy
i = 6% per year
100
94.34
$5.66 interest
$6.00 interest
Amount, $
50
0
1
Last
year
0 1 Time
Now Next
year
Figure 1–10
Equivalence of money at 6% per year interest.
EXAMPLE 1.12
Manufacturers make backup batteries for computer systems available to Batteries dealers
through privately owned distributorships. In general, batteries are stored throughout the year,
and a 5% cost increase is added each year to cover the inventory carrying charge for the distributorship
owner. Assume you own the City Center Batteries outlet. Make the calculations
necessary to show which of the following statements are true and which are false about battery
costs.
(a) The amount of $98 now is equivalent to a cost of $105.60 one year from now.
(b) A truck battery cost of $200 one year ago is equivalent to $205 now.
(c) A $38 cost now is equivalent to $39.90 one year from now.
(d) A $3000 cost now is equivalent to $2887.14 one year earlier.
(e) The carrying charge accumulated in 1 year on an investment of $20,000 worth of
batteries is $1000.
Solution
(a) Total amount accrued 98(1.05) $102.90 $105.60; therefore, it is false. Another
way to solve this is as follows: Required original cost is 105.601.05 $100.57 $98.
(b) Equivalent cost 1 year ago is 205.001.05 $195.24 $200; therefore, it is false.
(c) The cost 1 year from now is $38(1.05) $39.90; true.
(d) Cost now is 2887.14(1.05) $3031.50 $3000; false.
(e) The charge is 5% per year interest, or $20,000(0.05) $1000; true.
Comparison of alternative cash flow series requires the use of equivalence to determine when
the series are economically equal or if one is economically preferable to another. The keys to the
analysis are the interest rate and the timing of the cash flows. Example 1.13 demonstrates how
easy it is to be misled by the size and timing of cash flows.
EXAMPLE 1.13
Howard owns a small electronics repair shop. He wants to borrow $10,000 now and repay it
over the next 1 or 2 years. He believes that new diagnostic test equipment will allow him to
work on a wider variety of electronic items and increase his annual revenue. Howard received
2-year repayment options from banks A and B.

1.8 Simple and Compound Interest 21
Amount to pay, $ per year
Year Bank A Bank B
1 −5,378.05 −5,000.00
2 −5,378.05 −5,775.00
Total paid −10,756.10 −10,775.00
After reviewing these plans, Howard decided that he wants to repay the $10,000 after only
1 year based on the expected increased revenue. During a family conversation, Howard’s
brother-in-law offered to lend him the $10,000 now and take $10,600 after exactly 1 year.
Now Howard has three options and wonders which one to take. Which one is economically
the best?
Solution
The repayment plans for both banks are economically equivalent at the interest rate of 5% per
year. (This is determined by using computations that you will learn in Chapter 2.) Therefore,
Howard can choose either plan even though the bank B plan requires a slightly larger sum of
money over the 2 years.
The brother-in-law repayment plan requires a total of $600 in interest 1 year later plus the
principal of $10,000, which makes the interest rate 6% per year. Given the two 5% per year
options from the banks, this 6% plan should not be chosen as it is not economically better than
the other two. Even though the sum of money repaid is smaller, the timing of the cash flows
and the interest rate make it less desirable. The point here is that cash flows themselves, or
their sums, cannot be relied upon as the primary basis for an economic decision. The interest
rate, timing, and economic equivalence must be considered.
1.8 Simple and Compound Interest
The terms interest, interest period, and interest rate (introduced in Section 1.4) are useful in calculating
equivalent sums of money for one interest period in the past and one period in the future.
However, for more than one interest period, the terms simple interest and compound interest become
important.
Simple interest is calculated using the principal only, ignoring any interest accrued in preceding
interest periods. The total simple interest over several periods is computed as
Simple interest (principal)(number of periods)(interest rate) [1.7]
I Pni
where I is the amount of interest earned or paid and the interest rate i is expressed in decimal form.
EXAMPLE 1.14
GreenTree Financing lent an engineering company $100,000 to retrofit an environmentally
unfriendly building. The loan is for 3 years at 10% per year simple interest. How much money
will the firm repay at the end of 3 years?
Solution
The interest for each of the 3 years is
Interest per year $100,000(0.10) $10,000
Total interest for 3 years from Equation [1.7] is
Total interest $100,000(3)(0.10) $30,000

22 Chapter 1 Foundations of Engineering Economy
The amount due after 3 years is
Total due $100,000 30,000 $130,000
The interest accrued in the first year and in the second year does not earn interest. The interest
due each year is $10,000 calculated only on the $100,000 loan principal.
In most financial and economic analyses, we use compound interest calculations.
For compound interest, the interest accrued for each interest period is calculated on the
principal plus the total amount of interest accumulated in all previous periods. Thus,
compound interest means interest on top of interest.
Compound interest reflects the effect of the time value of money on the interest also. Now the
interest for one period is calculated as
Compound interest (principal all accrued interest)(interest rate) [1.8]
In mathematical terms, the interest I t for time period t may be calculated using the relation.
jt1
I t
(
P I J )( i) [1.9]
j1
EXAMPLE 1.15
Assume an engineering company borrows $100,000 at 10% per year compound interest and
will pay the principal and all the interest after 3 years. Compute the annual interest and total
amount due after 3 years. Graph the interest and total owed for each year, and compare with
the previous example that involved simple interest.
Solution
To include compounding of interest, the annual interest and total owed each year are calculated
by Equation [1.8].
Interest, year 1: 100,000(0.10) $10,000
Total due, year 1: 100,000 10,000 $110,000
Interest, year 2: 110,000(0.10) $11,000
Total due, year 2: 110,000 11,000 $121,000
Interest, year 3: 121,000(0.10) $12,100
Total due, year 3: 121,000 12,100 $133,100
The repayment plan requires no payment until year 3 when all interest and the principal, a total
of $133,100, are due. Figure 1–11 uses a cash flow diagram format to compare end-of-year
( a ) simple and ( b ) compound interest and total amounts owed. The differences due to compounding
are clear. An extra $133,100 – 130,000 $3100 in interest is due for the compounded
interest loan.
Note that while simple interest due each year is constant, the compounded interest due
grows geometrically. Due to this geometric growth of compound interest, the difference between
simple and compound interest accumulation increases rapidly as the time frame increases.
For example, if the loan is for 10 years, not 3, the extra paid for compounding interest
may be calculated to be $59,374.

1.8 Simple and Compound Interest 23
0
1 2 3 Year 0 1 2 3 Year
10
I I I
I is constant 10
I
11
12
I
I
I increases
geometrically
100
100
Amount owed ( $1000)
110
120
Amount owed ( $1000)
110
120
Arithmetic
increase
Geometric
increase
130
130
140
140
(a)
(b)
Figure 1–11
Interest I owed and total amount owed for ( a ) simple interest (Example 1.14) and ( b ) compound interest
(Example 1.15).
A more efficient way to calculate the total amount due after a number of years in Example 1.15
is to utilize the fact that compound interest increases geometrically. This allows us to skip the
year-by-year computation of interest. In this case, the total amount due at the end of each
year is
Year 1: $100,000(1.10)
1
$110,000
Year 2: $100,000(1.10) 2 $121,000
Year 3: $100,000(1.10) 3 $133,100
This allows future totals owed to be calculated directly without intermediate steps. The general
form of the equation is
Total due after n years principal(1 interest rate) n years [1.10]
P (1 i ) n
where i is expressed in decimal form. Equation [1.10] was applied above to obtain the
$133,100 due after 3 years. This fundamental relation will be used many times in the upcoming
chapters.
We can combine the concepts of interest rate, compound interest, and equivalence to demonstrate
that different loan repayment plans may be equivalent, but differ substantially in amounts
paid from one year to another and in the total repayment amount. This also shows that there are
many ways to take into account the time value of money.

24 Chapter 1 Foundations of Engineering Economy
EXAMPLE 1.16
Table 1–1 details four different loan repayment plans described below. Each plan repays a
$5000 loan in 5 years at 8% per year compound interest.
• Plan 1: Pay all at end. No interest or principal is paid until the end of year 5. Interest accumulates
each year on the total of principal and all accrued interest.
• Plan 2: Pay interest annually, principal repaid at end. The accrued interest is paid each
year, and the entire principal is repaid at the end of year 5.
• Plan 3: Pay interest and portion of principal annually. The accrued interest and one-fifth
of the principal (or $1000) are repaid each year. The outstanding loan balance decreases
each year, so the interest (column 2) for each year decreases.
• Plan 4: Pay equal amount of interest and principal. Equal payments are made each year
with a portion going toward principal repayment and the remainder covering the accrued
interest. Since the loan balance decreases at a rate slower than that in plan 3 due to the equal
end-of-year payments, the interest decreases, but at a slower rate.
TABLE 1–1 Different Repayment Schedules Over 5 Years for $5000 at 8% Per Year
Compound Interest
(1)
End of
Year
(2)
Interest Owed
for Year
Plan 1: Pay All at End
(3)
Total Owed at
End of Year
(4)
End-of-Year
Payment
(5)
Total Owed
After Payment
0 $5000.00
1 $400.00 $5400.00 — 5400.00
2 432.00 5832.00 — 5832.00
3 466.56 6298.56 — 6298.56
4 503.88 6802.44 — 6802.44
5 544.20 7346.64 $ – 7346.64
Total $ – 7346.64
Plan 2: Pay Interest Annually; Principal Repaid at End
0 $5000.00
1 $400.00 $5400.00 $400.00 5000.00
2 400.00 5400.00 400.00 5000.00
3 400.00 5400.00 400.00 5000.00
4 400.00 5400.00 400.00 5000.00
5 400.00 5400.00 – 5400.00
Total
Plan 3: Pay Interest and Portion of Principal Annually
$7000.00
0 $5000.00
1 $400.00 $5400.00 $1400.00 4000.00
2 320.00 4320.00 1320.00 3000.00
3 240.00 3240.00 1240.00 2000.00
4 160.00 2160.00 1160.00 1000.00
5 80.00 1080.00 – 1080.00
Total
Plan 4: Pay Equal Annual Amount of Interest and Principal
$6200.00
0 $5000.00
1 $400.00 $5400.00 $−1252.28 4147.72
2 331.82 4479.54 −1252.28 3227.25
3 258.18 3485.43 −1252.28 2233.15
4 178.65 2411.80 −1252.28 1159.52
5 92.76 1252.28 – 1252.28
Total
$−6261.40

1.9 Minimum Attractive Rate of Return 25
(a) Make a statement about the equivalence of each plan at 8% compound interest.
(b) Develop an 8% per year simple interest repayment plan for this loan using the same
approach as plan 2. Comment on the total amounts repaid for the two plans.
Solution
(a) The amounts of the annual payments are different for each repayment schedule, and
the total amounts repaid for most plans are different, even though each repayment
plan requires exactly 5 years. The difference in the total amounts repaid can be explained
by the time value of money and by the partial repayment of principal prior to
year 5.
A loan of $5000 at time 0 made at 8% per year compound interest is equivalent to each
of the following:
Plan 1 $7346.64 at the end of year 5
Plan 2 $400 per year for 4 years and $5400 at the end of year 5
Plan 3 Decreasing payments of interest and partial principal in years 1 ($1400)
through 5 ($1080)
Plan 4 $1252.28 per year for 5 years
An engineering economy study typically uses plan 4; interest is compounded, and a constant
amount is paid each period. This amount covers accrued interest and a partial
amount of principal repayment.
(b) The repayment schedule for 8% per year simple interest is detailed in Table 1–2. Since
the annual accrued interest of $400 is paid each year and the principal of $5000 is repaid
in year 5, the schedule is exactly the same as that for 8% per year compound interest, and
the total amount repaid is the same at $7000. In this unusual case, simple and compound
interest result in the same total repayment amount. Any deviation from this schedule will
cause the two plans and amounts to differ.
TABLE 1–2 A 5-Year Repayment Schedule of $5000 at 8% per Year Simple Interest
End of
Year
Interest Owed
for Year
Total Owed at
End of Year
End-of-Year
Payment
Total Owed
After Payment
0 $5000
1 $400 $5400 $400 5000
2 400 5400 400 5000
3 400 5400 400 5000
4 400 5400 400 5000
5 400 5400 – 5400 0
Total $7000
1.9 Minimum Attractive Rate of Return
For any investment to be profitable, the investor (corporate or individual) expects to receive more
money than the amount of capital invested. In other words, a fair rate of return, or return on investment,
must be realizable. The definition of ROR in Equation [1.4] is used in this discussion,
that is, amount earned divided by the principal.
Engineering alternatives are evaluated upon the prognosis that a reasonable ROR can be
expected. Therefore, some reasonable rate must be established for the selection criteria
(step 4) of the engineering economy study ( Figure 1–1 ).

26 Chapter 1 Foundations of Engineering Economy
Minimum Attractive Rate
of Return (MARR)
The Minimum Attractive Rate of Return (MARR) is a reasonable rate of return established
for the evaluation and selection of alternatives. A project is not economically viable unless
it is expected to return at least the MARR. MARR is also referred to as the hurdle rate,
cutoff rate, benchmark rate, and minimum acceptable rate of return.
Cost of capital
Figure 1–12 indicates the relations between different rate of return values. In the United
States, the current U.S. Treasury Bill return is sometimes used as the benchmark safe rate. The
MARR will always be higher than this, or a similar, safe rate. The MARR is not a rate that is
calculated as a ROR. The MARR is established by (financial) managers and is used as a criterion
against which an alternative’s ROR is measured, when making the accept/reject investment
decision.
To develop a foundation-level understanding of how a MARR value is established and used
to make investment decisions, we return to the term capital introduced in Section 1.1. Although
the MARR is used as a criterion to decide on investing in a project, the size of MARR is fundamentally
connected to how much it costs to obtain the needed capital funds. It always costs
money in the form of interest to raise capital. The interest, expressed as a percentage rate per
year, is called the cost of capital. As an example on a personal level, if you want to purchase a
new widescreen HDTV, but do not have sufficient money (capital), you could obtain a bank loan
for, say, a cost of capital of 9% per year and pay for the TV in cash now. Alternatively, you might
choose to use your credit card and pay off the balance on a monthly basis. This approach will
probably cost you at least 15% per year. Or, you could use funds from your savings account that
earns 5% per year and pay cash. This approach means that you also forgo future returns
from these funds. The 9%, 15%, and 5% rates are your cost of capital estimates to raise the
capital for the system by different methods of capital financing. In analogous ways, corporations
estimate the cost of capital from different sources to raise funds for engineering projects and
other types of projects.
Rate of return,
percent
Expected rate of return on
a new proposal
Range for the rate of return on
accepted proposals, if other
proposals were rejected
for some reason
All proposals must offer
at least MARR to
be considered
MARR
Rate of return on
“safe investment”
Figure 1–12
Size of MAAR relative to other rate of return values.

1.10 Introduction to Spreadsheet Use 27
In general, capital is developed in two ways—equity financing and debt financing. A combination
of these two is very common for most projects. Chapter 10 covers these in greater detail, but
a snapshot description follows.
Equity financing The corporation uses its own funds from cash on hand, stock sales, or retained
earnings. Individuals can use their own cash, savings, or investments. In the example above, using
money from the 5% savings account is equity financing.
Debt financing The corporation borrows from outside sources and repays the principal and interest
according to some schedule, much like the plans in Table 1–1. Sources of debt capital may be
bonds, loans, mortgages, venture capital pools, and many others. Individuals, too, can utilize debt
sources, such as the credit card (15% rate) and bank options (9% rate) described above.
Combinations of debt-equity financing mean that a weighted average cost of capital (WACC)
results. If the HDTV is purchased with 40% credit card money at 15% per year and 60% savings
account funds earning 5% per year, the weighted average cost of capital is 0.4(15) 0.6(5)
9% per year.
For a corporation, the established MARR used as a criterion to accept or reject an investment
alternative will usually be equal to or higher than the WACC that the corporation must bear to
obtain the necessary capital funds. So the inequality
ROR MARR WACC [1.11]
must be correct for an accepted project. Exceptions may be government-regulated requirements
(safety, security, environmental, legal, etc.), economically lucrative ventures expected to lead to
other opportunities, etc.
Often there are many alternatives that are expected to yield a ROR that exceeds the MARR as
indicated in Figure 1–12 , but there may not be sufficient capital available for all, or the project’s
risk may be estimated as too high to take the investment chance. Therefore, new projects that are
undertaken usually have an expected return at least as great as the return on another alternative
that is not funded. The expected rate of return on the unfunded project is called the opportunity
cost.
The opportunity cost is the rate of return of a forgone opportunity caused by the inability to
pursue a project. Numerically, it is the largest rate of return of all the projects not accepted
(forgone) due to the lack of capital funds or other resources. When no specific MARR is
established, the de facto MARR is the opportunity cost, i.e., the ROR of the first project not
undertaken due to unavailability of capital funds.
Opportunity cost
As an illustration of opportunity cost, refer to Figure 1–12 and assume a MARR of 12% per
year. Further, assume that a proposal, call it A, with an expected ROR 13% is not funded due
to a lack of capital. Meanwhile, proposal B has a ROR 14.5% and is funded from available
capital. Since proposal A is not undertaken due to the lack of capital, its estimated ROR of 13%
is the opportunity cost; that is, the opportunity to make an additional 13% return is forgone.
1.10 Introduction to Spreadsheet Use
The functions on a computer spreadsheet can greatly reduce the amount of hand work for equivalency
computations involving compound interest and the terms P , F , A , i , and n . The use of a
calculator to solve most simple problems is preferred by many students and professors as described
in Appendix D. However, as cash flow series become more complex, the spreadsheet
offers a good alternative. Microsoft Excel is used throughout this book because it is readily
available and easy to use. Appendix A is a primer on using spreadsheets and Excel. The functions
used in engineering economy are described there in detail, with explanations of all the

28 Chapter 1 Foundations of Engineering Economy
parameters. Appendix A also includes a section on spreadsheet layout that is useful when the
economic analysis is presented to someone else—a coworker, a boss, or a professor.
A total of seven Excel functions can perform most of the fundamental engineering economy
calculations. The functions are great supplemental tools, but they do not replace the understanding
of engineering economy relations, assumptions, and techniques. Using the symbols P , F , A ,
i , and n defined in the previous section, the functions most used in engineering economic analysis
are formulated as follows.
To find the present value P : PV( i%, n, A, F )
To find the future value F : FV( i%, n, A, P )
To find the equal, periodic value A : PMT( i%, n, P, F )
To find the number of periods n : NPER( i%, A, P, F )
To find the compound interest rate i : RATE( n, A, P, F )
To find the compound interest rate i : IRR(first_cell:last_cell)
To find the present value P of any series: NPV( i %, second_cell:last_cell) first_cell
If some of the parameters don’t apply to a particular problem, they can be omitted and zero is
assumed. For readability, spaces can be inserted between parameters within parentheses. If the
parameter omitted is an interior one, the comma must be entered. The last two functions require
that a series of numbers be entered into contiguous spreadsheet cells, but the first five can be used
with no supporting data. In all cases, the function must be preceded by an equals sign () in the
cell where the answer is to be displayed.
To understand how the spreadsheet functions work, look back at Example 1.6 a , where the
equivalent annual amount A is unknown, as indicated by A ?. (In Chapter 2, we learn how
engineering economy factors calculate A , given P , i , and n .) To find A using a spreadsheet
function, simply enter the PMT function PMT(5%,5,5000). Figure 1–13 is a screen image
of a spreadsheet with the PMT function entered into cell B4. The answer ($1154.87) is displayed.
The answer may appear in red and in parentheses, or with a minus sign on your screen
to indicate a negative amount from the perspective of a reduction in the account balance. The
right side of Figure 1–13 presents the solution to Example 1.6 b. The future value F is determined
by using the FV function. The FV function appears in the formula bar; and many examples
throughout this text will include cell tags, as shown here, to indicate the format of
important entries.
The following example demonstrates the use of a spreadsheet to develop relations (not
built-in functions) to calculate interest and cash flows. Once set up, the spreadsheet can be
used to perform sensitivity analysis for estimates that are subject to change. We will illustrate
the use of spreadsheets throughout the chapters. ( Note: The spreadsheet examples may
be omitted, if spreadsheets are not used in the course. A solution by hand is included in virtually
all examples.)
PMT(5%,5,5000)
FV(7%,3,,5000)
Figure 1–13
Use of spreadsheet functions PMT and FV, Example 1.6.

1.10 Introduction to Spreadsheet Use 29
EXAMPLE 1.17
A Japan-based architectural firm has asked a United States–based software engineering group
to infuse GPS sensing capability via satellite into monitoring software for high-rise structures
in order to detect greater than expected horizontal movements. This software could be very
beneficial as an advance warning of serious tremors in earthquake-prone areas in Japan and the
United States. The inclusion of accurate GPS data is estimated to increase annual revenue over
that for the current software system by $200,000 for each of the next 2 years, and by $300,000
for each of years 3 and 4. The planning horizon is only 4 years due to the rapid advances made
internationally in building-monitoring software. Develop spreadsheets to answer the questions
below.
(a) Determine the total interest and total revenue after 4 years, using a compound rate of
r eturn of 8% per year.
(b) Repeat part (a) if estimated revenue increases from $300,000 to $600,000 in years 3 and 4.
(c) Repeat part ( a ) if inflation is estimated to be 4% per year. This will decrease the real rate
of return from 8% to 3.85% per year (Chapter 14 shows why).
Solution by Spreadsheet
Refer to Figure 1–14 a to d for the solutions. All the spreadsheets contain the same information,
but some cell values are altered as required by the question. (Actually, all the questions can be
answered on one spreadsheet by changing the numbers. Separate spreadsheets are shown here
for explanation purposes only.)
The Excel functions are constructed with reference to the cells, not the values themselves,
so that sensitivity analysis can be performed without function changes. This approach
treats the value in a cell as a global variable for the spreadsheet. For example, the
8% rate in cell B2 will be referenced in all functions as B2, not 8%. Thus, a change in the
rate requires only one alteration in the cell B2 entry, not in every relation where 8% is used.
See Appendix A for additional information about using cell referencing and building
spreadsheet relations.
(a) Figure 1–14 a shows the results, and Figure 1–14 b presents all spreadsheet relations for
estimated interest and revenue (yearly in columns C and E, cumulative in columns D
and F). As an illustration, for year 3 the interest I 3 and revenue plus interest R 3 are
I 3 (cumulative revenue through year 2)(rate of return)
$416,000(0.08)
$33,280
R 3 revenue in year 3 I 3
$300,000 33,280
$333,280
The detailed relations shown in Figure 1–14 b calculate these values in cells C8 and E8.
Cell C8 relation for I 3 : F7*B2
Cell E8 relation for CF 3 : B8 C8
The equivalent amount after 4 years is $1,109,022, which is comprised of $1,000,000 in
total revenue and $109,022 in interest compounded at 8% per year. The shaded cells in
Figure 1–14 a and b indicate that the sum of the annual values and the last entry in the cumulative
columns must be equal.
(b) To determine the effect of increasing estimated revenue for years 3 and 4 to $600,000,
use the same spreadsheet and change the entries in cells B8 and B9 as shown in
Figure 1–14 c . Total interest increases 22%, or $24,000, from $109,222 to $133,222.
(c) Figure 1–14 d shows the effect of changing the original i value from 8% to an inflationadjusted
rate of 3.85% in cell B2 on the first spreadsheet. [Remember to return to the
$300,000 revenue estimates for years 3 and 4 after working part ( b ).] Inflation has now
reduced total interest by 53% from $109,222 to $51,247, as shown in cell C10.

30 Chapter 1 Foundations of Engineering Economy
(a) Total interest and revenue for base case, year 4
(b) Spreadsheet relations for base case
Revenue changed
(c) Totals with increased revenue in years 3 and 4
Rate of
return
changed
(d) Totals with inflation of 4% per year considered
Figure 1–14
Spreadsheet solutions with sensitivity analysis, Example 1.17 a to c .
Comment
Later we will learn how to utilize the NPV and FV Excel financial functions to obtain the same
answers determined in Figure 1–14 , where we developed each basic relation.
When you are working with an Excel spreadsheet, it is possible to display all of the entries
and functions on the screen as shown in Figure 1–14 b by simultaneously touching the
and < `> keys, which may be in the upper left of the keyboard on the key with .

Problems 31
CHAPTER SUMMARY
Engineering economy is the application of economic factors and criteria to evaluate alternatives,
considering the time value of money. The engineering economy study involves computing a
specific economic measure of worth for estimated cash flows over a specific period of time.
The concept of equivalence helps in understanding how different sums of money at different
times are equal in economic terms. The differences between simple interest (based on principal
only) and compound interest (based on principal and interest upon interest) have been described
in formulas, tables, and graphs. This power of compounding is very noticeable, especially over
extended periods of time, and for larger sums of money.
The MARR is a reasonable rate of return established as a hurdle rate to determine if an alternative
is economically viable. The MARR is always higher than the return from a safe investment
and the cost to acquire needed capital.
Also, we learned a lot about cash flows:
End-of-year convention for cash flow location
Net cash flow computation
Different perspectives in determining the cash flow sign
Construction of a cash flow diagram
Difficulties with estimating future cash flows accurately
PROBLEMS
Basic Concepts
1.1 List the four essential elements involved in decision
making in engineering economic analysis.
1.2 What is meant by ( a ) limited capital funds and
( b ) sensitivity analysis?
1.3 List three measures of worth that are used in engineering
economic analysis.
1.4 Identify the following factors as either economic
(tangible) or noneconomic (intangible): first cost,
leadership, taxes, salvage value, morale, dependability,
inflation, profit, acceptance, ethics, interest
rate.
Ethics
1.5 Stefanie is a design engineer with an international
railroad locomotive manufacturing company in
Illinois. Management wants to return some of the
engineering design work to the United States
rather than export all of it to India, where the primary
design work has been accomplished for the
last decade. This transfer will employ more people
locally and could improve the economic conditions
for families in and around Illinois.
Stefanie and her design team were selected as a
test case to determine the quality and speed of the
design work they could demonstrate on a more
fuel-efficient diesel locomotive. Neither she nor
any of her team members have done such a significant
design job, because their jobs had previously
entailed only the interface with the subcontracted
engineers in India. One of her team members had a
great design idea on a key element that will improve
fuel efficiency by approximately 15%. She
told Stefanie it came from one of the Indiangenerated
documents, but that it would probably be
okay for the team to use it and remain silent as to its
origin, since it was quite clear the U.S. management
was about to cancel the foreign contract.
Although reluctant at first, Stefanie did go forward
with a design that included the efficiency improvement,
and no mention of the origin of the idea was
made at the time of the oral presentation or documentation
delivery. As a result, the Indian contract
was canceled and full design responsibility was
transferred to Stefanie’s group.
Consult the NSPE Code of Ethics for Engineers
(Appendix C) and identify sections that are points
of concern about Stefanie’s decisions and actions.
1.6 Consider the common moral precept that stealing
is wrong. Hector is with a group of friends in a
local supermarket. One of Hector’s buddies takes a
high-energy drink from a six-pack on the shelf,
opens it, drinks it, and returns the empty can to the
package, with no intention of paying for it. He then
invites the others to do the same, saying, “It’s only

32 Chapter 1 Foundations of Engineering Economy
one drink. Others do it all the time.” All the others,
except Hector, have now consumed a drink of their
choice. Personally, Hector believes this is a form
of stealing. State three actions that Hector can
take, and evaluate them from the personal moral
perspective.
1.7 While going to work this morning off site from his
office, an engineer accidently ran a stop sign and
was in a car accident that resulted in the death of a
5-year-old child. He has a strong belief in the universal
moral that it is wrong to do serious harm to
another person. Explain the conflict that can arise
for him between the universal moral and his personal
moral about doing serious harm, given the
accident was deemed his fault.
1.8 Claude is a fourth-year engineering university student
who has just been informed by his instructor
that he made a very low grade on his Spanish language
final test for the year. Although he had a
passing score prior to the final, his final grade was
so low that he has now flunked the entire year and
will likely have to extend his graduation another
semester or two.
Throughout the year, Claude, who hated the
course and his instructor, has copied homework,
cheated on tests, and never seriously studied for
anything in the course. He did realize during the
semester that he was doing something that even
he consi dered wrong morally and ethically. He
knew he had done badly on the final. The classroom
was reconfigured for the final exam in a way
that he could not get any answers from classmates,
and cell phones were collected prior to the exam,
thus removing texting possibilities to friends outside
the classroom who might help him on the
final exam. Claude is now face to face with the
instructor in her office. The question to Claude is,
“What have you been doing throughout this year
to make passing scores repeatedly, but demonstrate
such a poor command of Spanish on the
final exam?”
From an ethical viewpoint, what options does
Claude have in his answer to this question? Also,
discuss some of the possible effects that this experience
may have upon Claude’s future actions and
moral dilemmas.
Interest Rate and Rate of Return
1.9 RKI Instruments borrowed $3,500,000 from a private
equity firm for expansion of its manufacturing
facility for making carbon monoxide monitors/
controllers. The company repaid the loan after
1 year with a single payment of $3,885,000. What
was the interest rate on the loan?
1.10 Emerson Processing borrowed $900,000 for installing
energy-efficient lighting and safety equipment
in its La Grange manufacturing facility. The
terms of the loan were such that the company could
pay interest only at the end of each year for up to 5
years, after which the company would have to pay
the entire amount due. If the interest rate on the
loan was 12% per year and the company paid only
the interest for 4 years, determine the following:
(a) The amount of each of the four interest
payments
(b) The amount of the final payment at the end of
year 5
1.11 Which of the following 1-year investments has the
highest rate of return?
(a) $12,500 that yields $1125 in interest,
(b) $56,000 that yields $6160 in interest, or
(c) $95,000 that yields $7600 in interest .
1.12 A new engineering graduate who started a consulting
business borrowed money for 1 year to furnish
the office. The amount of the loan was $23,800,
and it had an interest rate of 10% per year. However,
because the new graduate had not built up a
credit history, the bank made him buy loan-default
insurance that cost 5% of the loan amount. In addition,
the bank charged a loan setup fee of $300.
What was the effective interest rate the engineer
paid for the loan?
1.13 When the inflation rate is expected to be 8% per
year, what is the market interest rate likely to be?
Terms and Symbols
1.14 The symbol P represents an amount of money at a
time designated as present. The following symbols
also represent a present amount of money and require
similar calculations. Explain what each symbol
stands for: PW, PV, NPV, DCF, and CC .
1.15 Identify the four engineering economy symbols
and their values from the following problem statement.
Use a question mark with the symbol whose
value is to be determined.
Thompson Mechanical Products is planning to
set aside $150,000 now for possibly replacing its
large synchronous refiner motors whenever it becomes
necessary. If the replacement is not needed
for 7 years, how much will the company have in its
investment set-aside account, provided it achieves
a rate of return of 11% per year?
1.16 Identify the four engineering economy symbols and
their values from the following problem statement.

Problems 33
Use a question mark with the symbol whose value is
to be determined.
Atlas Long-Haul Transportation is considering
installing Valutemp temperature loggers in all
of its refrigerated trucks for monitoring temperatures
during transit. If the systems will reduce
insurance claims by $100,000 two years from
now, how much should the company be willing to
spend now, if it uses an interest rate of 12% per
year?
1.17 Identify the four engineering economy symbols
and their values from the following problem statement.
Use a question mark with the symbol whose
value is to be determined.
A green algae, Chlamydomonas reinhardtii,
can produce hydrogen when temporarily deprived
of sulfur for up to 2 days at a time. A small
company needs to purchase equipment costing
$3.4 million to commercialize the process. If the
company wants to earn a rate of return of 10% per
year and recover its investment in 8 years, what
must be the net value of the hydrogen produced
each year?
1.18 Identify the four engineering economy symbols
and their values from the following problem statement.
Use a question mark with the symbol whose
value is to be determined.
Vision Technologies, Inc., is a small company
that uses ultra-wideband technology to develop
devices that can detect objects (including people)
inside of buildings, behind walls, or below ground.
The company expects to spend $100,000 per year
for labor and $125,000 per year for supplies before
a product can be marketed. At an interest rate of
15% per year, what is the total equivalent future
amount of the company’s expenses at the end of
3 years?
Cash Flows
1.19 What is meant by end-of-period convention?
1.20 Identify the following as cash inflows or outflows
to commercial air carriers: fuel cost, pension plan
contributions, fares, maintenance, freight revenue,
cargo revenue, extra-bag charges, water and
sodas, advertising, landing fees, seat preference
fees.
1.21 Many credit unions use semiannual interest periods
to pay interest on customer savings accounts. For a
credit union that uses June 30 and December 31 as
its semiannual interest periods, determine the endof-period
amounts that will be recorded for the deposits
shown in the table.
Month Deposit, $
Jan 50
Feb 70
Mar —
Apr 120
May 20
June —
July 150
Aug 90
Sept —
Oct —
Nov 40
Dec 110
1.22 For a company that uses a year as its interest period,
determine the net cash fl ow that will be recorded
at the end of the year from the cash flows
shown.
Month
Receipts,
$1000
Disbursements,
$1000
Jan 500 300
Feb 800 500
Mar 200 400
Apr 120 400
May 600 500
June 900 600
July 800 300
Aug 700 300
Sept 900 500
Oct 500 400
Nov 400 400
Dec 1800 700
1.23 Construct a cash flow diagram for the following
cash flows: $25,000 outflow at time 0, $9000 per
year inflow in years 1 through 5 at an interest rate
of 10% per year, and an unknown future amount in
year 5.
1.24 Construct a cash flow diagram to find the present
worth in year 0 at an interest rate of 15% per year
for the following situation.
Year Cash Flow, $
0 19,000
1–4 8,100
1.25 Construct a cash flow diagram that represents the
amount of money that will be accumulated in
15 years from an investment of $40,000 now at an
interest rate of 8% per year.
Equivalence
1.26 At an interest rate of 15% per year, an investment
of $100,000 one year ago is equivalent to how
much now?

34 Chapter 1 Foundations of Engineering Economy
1.27 During a recession, the price of goods and services
goes down because of low demand. A company
that makes Ethernet adapters is planning to expand
its production facility at a cost of $1,000,000 one
year from now. However, a contractor who needs
work has offered to do the job for $790,000 if the
company will do the expansion now instead of
1 year from now. If the interest rate is 15% per year,
how much of a discount is the company getting?
1.28 As a principal in the consulting firm where you
have worked for 20 years, you have accumulated
5000 shares of company stock. One year ago, each
share of stock was worth $40. The company has
offered to buy back your shares for $225,000. At
what interest rate would the firm’s offer be equivalent
to the worth of the stock last year?
1.29 A design/build engineering company that usually
gives year-end bonuses in the amount of $8000 to
each of its engineers is having cash flow problems.
The company said that although it could not
give bonuses this year, it would give each engineer
two bonuses next year, the regular one of
$8000 plus an amount equivalent to the $8000
that each engineer should have gotten this year.
If the interest rate is 8% per year, what will be the
total amount of bonus money the engineers should
get next year?
1.30 University tuition and fees can be paid by using
one of two plans.
Early-bird: Pay total amount due 1 year in
advance and get a 10% discount.
On-time: Pay total amount due when classes start.
The cost of tuition and fees is $10,000 per year.
(a) How much is paid in the early-bird plan?
(b) What is the equivalent amount of the savings
compared to the on-time payment at the time
that the on-time payment is made?
Simple and Compound Interest
1.31 If a company sets aside $1,000,000 now into a
contingency fund, how much will the company
have in 2 years, if it does not use any of the money
and the account grows at a rate of 10% per year?
1.32 Iselt Welding has extra funds to invest for future
capital expansion. If the selected investment pays
simple interest, what interest rate would be
required for the amount to grow from $60,000 to
$90,000 in 5 years?
1.33 To finance a new product line, a company that
makes high-temperature ball bearings borrowed
$1.8 million at 10% per year interest. If the company
repaid the loan in a lump sum amount after
2 years, what was ( a ) the amount of the payment
and ( b ) the amount of interest?
1.34 Because market interest rates were near all-time
lows at 4% per year, a hand tool company decided
to call (i.e., pay off ) the high-interest bonds that it
issued 3 years ago. If the interest rate on the bonds
was 9% per year, how much does the company
have to pay the bond holders? The face value
(principal) of the bonds is $6,000,000.
1.35 A solid waste disposal company borrowed money
at 10% per year interest to purchase new haulers
and other equipment needed at the companyowned
landfill site. If the company got the loan
2 years ago and paid it off with a single payment of
$4,600,000, what was the principal amount P of
the loan?
1.36 If interest is compounded at 20% per year, how
long will it take for $50,000 to accumulate to
$86,400?
1.37 To make CDs look more attractive than they really
are, some banks advertise that their rates are higher
than their competitors’ rates; however, the fine
print says that the rate is a simple interest rate. If a
person deposits $10,000 at 10% per year simple
interest, what compound interest rate would yield
the same amount of money in 3 years?
MARR and Opportunity Cost
1.38 Give three other names for minimum attractive
rate of return.
1.39 Identify the following as either equity or debt financing:
bonds, stock sales, retained earnings,
venture capital, short-term loan, capital advance
from friend, cash on hand, credit card, home equity
loan.
1.40 What is the weighted average cost of capital for a
corporation that finances an expansion project
using 30% retained earnings and 70% venture capital?
Assume the interest rates are 8% for the equity
financing and 13% for the debt financing.
1.41 Managers from different departments in Zenith
Trading, a large multinational corporation, have offered
six projects for consideration by the corporate
office. A staff member for the chief financial officer
used key words to identify the projects and then
listed them in order of projected rate of return as
shown below. If the company wants to grow rapidly
through high leverage and uses only 10%
equity financing that has a cost of equity capital of
9% and 90% debt financing with a cost of debt

Additional Problems and FE Exam Review Questions 35
capital of 16%, which projects should the company
undertake?
Project ID
Projected ROR,
% per year
Inventory 30
Technology 28.4
Warehouse 19
Products 13.1
Energy 9.6
Shipping 8.2
Spreadsheet Functions
1.42 State the purpose for each of the following built-in
spreadsheet functions.
( a) PV( i% , n , A , F )
( b) FV( i %, n , A , P )
( c) RATE( n , A , P , F )
( d) IRR(first_cell:last_cell)
( e) PMT( i %, n , P , F )
( f ) NPER( i %, A , P , F )
1.43 What are the values of the engineering economy
symbols P , F , A , i , and n in the following functions?
Use a question mark for the symbol that is to be
determined.
(a) NPER(8%,1500,8000,2000)
( b ) FV(7%,102000,9000)
(c) RATE(10,1000,12000,2000)
(d ) PMT(11%,20,,14000)
(e) PV(8%,15,1000,800)
1.44 Write the engineering economy symbol that corresponds
to each of the following spreadsheet
functions.
( a ) PMT ( b ) FV ( c ) NPER ( d ) PV ( e ) IRR
1.45 In a built-in spreadsheet function, if a certain parameter
is not present, ( a ) under what circumstances
can it be left blank and ( b ) when must a
comma be entered in its place?
1.46 Sheryl and Marcelly both invest $1000 at 10% per
year for 4 years. Sheryl receives simple interest and
Marcelly gets compound interest. Use a spreadsheet
and cell reference formats to develop relations that
show a total of $64 more interest for Marcelly at the
end of the 4 years. Assume no withdrawals or further
deposits are made during the 4 years.
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
1.47 The concept that different sums of money at different
points in time can be said to be equal to each
other is known as:
( a) Evaluation criterion
( b) Equivalence
( c) Cash flow
( d ) Intangible factors
1.48 The evaluation criterion that is usually used in an
economic analysis is:
( a) Time to completion
( b) Technical feasibility
( c) Sustainability
( d ) Financial units (dollars or other currency)
1.49 All of the following are examples of cash outflows,
except :
( a) Asset salvage value
( b) Income taxes
( c) Operating cost of asset
( d ) First cost of asset
1.50 In most engineering economy studies, the best alternative
is the one that:
( a) Will last the longest time
( b) Is most politically correct
( c) Is easiest to implement
( d ) Has the lowest cost
1.51 The following annual maintenance and operation
(M&O) costs for a piece of equipment were collected
over a 5-year period: $12,300, $8900,
$9200, $11,000, and $12,100. The average is
$10,700. In conducting a sensitivity analysis, the
most reasonable range of costs to use (i.e., percent
from the average) is:
( a ) 5% ( b ) 11% ( c ) 17% ( d ) 25%
1.52 At an interest rate of 10% per year, the equivalent
amount of $10,000 one year ago is closest to:
( a ) $8264 ( b ) $9091 ( c ) $11,000 ( d ) $12,000
1.53 Assume that you and your best friend each have
$1000 to invest. You invest your money in a fund
that pays 10% per year compound interest. Your
friend invests her money at a bank that pays 10%
per year simple interest. At the end of 1 year, the
difference in the total amount for each of you is:
( a) You have $10 more than she does
( b) You have $100 more than she does
( c) You both have the same amount of money
( d ) She has $10 more than you do
1.54 The time it would take for a given sum of money to
double at 4% per year simple interest is closest to:
( a ) 30 years ( b ) 25 years
( c ) 20 years ( d ) 10 years

36 Chapter 1 Foundations of Engineering Economy
1.55 All of the following are examples of equity financing,
except :
(a) Mortgage
(b) Money from savings
(c) Cash on hand
(d ) Retained earnings
1.56 To finance a new project costing $30 million, a
company borrowed $21 million at 16% per year
interest and used retained earnings valued at 12%
per year for the remainder of the investment. The
company’s weighted average cost of capital for the
project was closest to:
( a ) 12.5% ( b ) 13.6% ( c ) 14.8% ( d ) 15.6%
CASE STUDY
RENEWABLE ENERGY SOURCES FOR ELECTRICITY GENERATION
Background
Pedernales Electric Cooperative (PEC) is the largest
member-owned electric co-op in the United States with over
232,000 meters in 12 Central Texas counties. PEC has a capacity
of approximately 1300 MW (megawatts) of power, of
which 277 MW, or about 21%, is from renewable sources.
The latest addition is 60 MW of power from a wind farm in
south Texas close to the city of Corpus Christi. A constant
question is how much of PEC’s generation capacity should be
from renewable sources, especially given the environmental
issues with coal-generated electricity and the rising costs of
hydrocarbon fuels.
Wind and nuclear sources are the current consideration for
the PEC leadership as Texas is increasing its generation by
nuclear power and the state is the national leader in wind
farm–produced electricity.
Consider yourself a member of the board of directors of
PEC. You are an engineer who has been newly elected by the
PEC membership to serve a 3-year term as a director-at-large.
As such, you do not represent a specific district within the
entire service area; all other directors do represent a specific
district. You have many questions about the operations of
PEC, plus you are interested in the economic and societal
benefits of pursuing more renewable source generation
capacity.
Information
Here are some data that you have obtained. The information
is sketchy, as this point, and the numbers are very approximate.
Electricity generation cost estimates are national
in scope, not PEC-specific, and are provided in cents per
kilowatt-hour (¢/kWh).
Generation Cost, ¢/kWh
Fuel Source Likely Range Reasonable Average
Coal 4 to 9 7.4
Natural gas 4 to 10.5 8.6
Wind 4.8 to 9.1 8.2
Solar 4.5 to 15.5 8.8
National average cost of electricity to residential customers:
11¢/kWh
PEC average cost to residential customers: 10.27 ¢/kWh
(from primary sources) and 10.92 ¢/kWh (renewable sources)
Expected life of a generation facility: 20 to 40 years (it is
likely closer to 20 than 40)
Time to construct a facility: 2 to 5 years
Capital cost to build a generation facility: $900 to $1500
per kW
You have also learned that the PEC staff uses the wellrecognized
levelized energy cost (LEC) method to determine
the price of electricity that must be charged to customers to
break even. The formula takes into account the capital cost of
the generation facilities, the cost of capital of borrowed
money, annual maintenance and operation (M&O) costs, and
the expected life of the facility. The LEC formula, expressed
in dollars per kWh for ( t 1, 2, ... , n ), is
tn
P t A t C
—————— t
t1 (1 i)
LEC
t
———————
tn
E
——— t
t1 (1 i) t
where P t capital investments made in year t
A t annual maintenance and operating (M&O) costs
for year t
C t fuel costs for year t
E t amount of electricity generated in year t
n expected life of facility
i discount rate (cost of capital)
Case Study Exercises
1. If you wanted to know more about the new arrangement
with the wind farm in south Texas for the additional
60 MW per year, what types of questions would
you ask of a staff member in your first meeting with
him or her?
2. Much of the current generation capacity of PEC facilities
utilizes coal and natural gas as the primary fuel source.
What about the ethical aspects of the government’s allowance
for these plants to continue polluting the atmosphere
with the emissions that may cause health problems for
citizens and further the effects of global warming? What
types of regulations, if any, should be developed for PEC
(and other generators) to follow in the future?

Case Study 37
3. You developed an interest in the LEC relation and
the publicized cost of electricity of 10.27¢/kWh for
this year. You wonder if the addition of 60 MW of
wind-sourced electricity will make any difference in
the LEC value for this next year. You did learn the
following:
This is year t 11 for LEC computation purposes
n 25 years
i 5% per year
E 11 5.052 billion kWh
LEC last year was 10.22 ¢/kWh (last year’s breakeven
cost to customers)
From these sketchy data, can you determine the value of unknowns
in the LEC relation for this year? Is it possible to
determine if the wind farm addition of 60 MW makes any
difference in the electricity rate charged to customers? If not,
what additional information is necessary to determine the
LEC with the wind source included?
CASE STUDY
REFRIGERATOR SHELLS
Background
Large refrigerator manufacturers such as Whirlpool, General
Electric, Frigidaire, and others may subcontract the molding of
their plastic liners and door panels. One prime national subcontractor
is Innovations Plastics. Because of improvements in mechanical
properties, the molded plastic can sustain increased vertical
and horizontal loading, thus significantly reducing the need
for attached metal anchors for some shelving. However, improved
molding equipment is needed to enter this market now.
The company president wants a recommendation on whether
Innovations should offer the new technology to the major manufacturers
and an estimate of the necessary capital investment
to enter this market.
You work as an engineer for Innovations. At this stage,
you are not expected to perform a complete engineering economic
analysis, for not enough information is available. You
are asked to formulate reasonable alternatives, determine
what data and estimates are needed for each one, and ascertain
what criteria (economic and noneconomic) should be
utilized to make the final decision.
Information
Some information useful at this time is as follows:
• The technology and equipment are expected to last about
10 years before new methods are developed.
• Inflation and income taxes will not be considered in the
analysis.
• The expected returns on capital investment used for the
last three new technology projects were compound rates of
15%, 5%, and 18%. The 5% rate was the criterion for
enhancing an employee-safety system on an existing
chemical-mixing process.
• Equity capital financing beyond $5 million is not possible.
The amount of debt financing and its cost are unknown.
• Annual operating costs have been averaging 8% of first
cost for major equipment.
• Increased annual training costs and salary requirements
for handling the new plastics and operating new equipment
can range from $800,000 to $1.2 million.
There are two manufacturers working on the new- generation
equipment. You label these options as alternatives A and B.
Case Study Exercises
1. Use the first four steps of the decision-making process
to generally describe the alternatives and identify what
economic-related estimates you will need to complete
an engineering economy analysis for the president.
2. Identify any noneconomic factors and criteria to be considered
in making the alternative selection.
3. During your inquiries about alternative B from its manufacturer,
you learn that this company has already produced
a prototype molding machine and has sold it to a company
in Germany for $3 million (U.S. dollars). Upon inquiry,
you further discover that the German company already
has unused capacity on the equipment for manufacturing
plastic shells. The company is willing to sell time on the
equipment to Innovations immediately to produce its own
shells for U.S. delivery. This could allow immediate market
entry into the United States. Consider this as alternative
C, and develop the estimates necessary to evaluate C
at the same time as alternatives A and B.

CHAPTER 2
Factors: How
Time and
Interest Affect
Money
LEARNING OUTCOMES
Purpose: Derive and use the engineering economy factors to account for the time value of money.
SECTION TOPIC LEARNING OUTCOME
2.1 FP and PF factors • Derive and use factors for single amounts—
compound amount (FP) and present worth (PF)
factor.
2.2 PA and AP factors • Derive and use factors for uniform series—present
worth (PA) and capital recovery (AP) factors.
2.3 FA and AF factors • Derive and use factors for uniform series—
compound amount (FA) and sinking fund (AF)
factors.
2.4 Factor values • Use linear interpolation in factor tables or
spreadsheet functions to determine factor values.
2.5 Arithmetic gradient • Use the present worth (PG) and uniform annual
series (AG) factors for arithmetic gradients.
2.6 Geometric gradient • Use the geometric gradient series factor (PA,g)
to find present worth.
2.7 Find i or n • Use equivalence relations to determine i (interest
rate or rate of return) or n for a cash flow series.

T
he cash flow is fundamental to every economic study. Cash flows occur in many
configurations and amounts—isolated single values, series that are uniform, and
series that increase or decrease by constant amounts or constant percentages.
This chapter develops derivations for all the commonly used engineering economy factors
that take the time value of money into account.
The application of factors is illustrated using their mathematical forms and a standard notation
format. Spreadsheet functions are used in order to rapidly work with cash flow series
and to perform sensitivity analysis.
If the derivation and use of factors are not covered in the course, alternate ways to perform
time value of money calculations are summarized in Appendix D.
The Cement Factory Case: Votorantim
Cimentos North America, Inc., is a subsidiary
of a Brazil-based company that
recently announced plans to develop a
new cement factory in Houston County
in the state of Georgia. The plant will
be called Houston American Cement, or
HAC. The location is ideal for cement
making because of the large deposit of
limestone in the area.
The plant investment, expected
to amount to $200 million, has been
planned for 2012; however, it is currently
delayed due to the economic downturn
in construction. When the plant is completed
and operating at full capacity,
PE
based upon the projected needs and
cost per metric ton, it is possible that
the plant could generate as much as
$50,000,000 annually in revenue. All
analysis will use a planning horizon of
5 years commencing when the plant
begins operation.
This case is used in the following
topics (and sections) of this chapter:
Single-amount factors (2.1)
Uniform series factors (2.2 and 2.3)
Arithmetic gradient factors (2.5)
Geometric gradient factors (2.6)
Determining unknown n values (2.7)
2.1 Single-Amount Factors ( F P and P F )
The most fundamental factor in engineering economy is the one that determines the amount
of money F accumulated after n years (or periods) from a single present worth P, with interest
compounded one time per year (or period). Recall that compound interest refers to interest paid
on top of interest. Therefore, if an amount P is invested at time t 0, the amount F 1 accumulated
1 year hence at an interest rate of i percent per year will be
F 1 P Pi
P (1 i )
where the interest rate is expressed in decimal form. At the end of the second year, the amount
accumulated F 2 is the amount after year 1 plus the interest from the end of year 1 to the end of
year 2 on the entire F 1 .
The amount F 2 can be expressed as
F 2 F 1 F 1 i
P (1 i ) P (1 i ) i [2.1]
F 2 P (1 i i i 2 )
P (1 2 i i 2 )
P (1 i )
2
Similarly, the amount of money accumulated at the end of year 3, using Equation [2.1], will be
F 3 F 2 F 2 i

40 Chapter 2 Factors: How Time and Interest Affect Money
F =?
F = given
i = given
i = given
0 1 2 n – 2 n – 1 n
0 1 2 n – 2 n – 1 n
P = given
(a)
P =?
Figure 2–1
Cash flow diagrams for single-payment factors: (a) find F, given P, and (b) find P, given F.
(b)
Substituting P (1 i ) 2 for F 2 and simplifying, we get
F 3 P (1 i ) 3
From the preceding values, it is evident by mathematical induction that the formula can be generalized
for n years. To find F , given P ,
F P(1 i) n [2.2]
The factor (1 i ) n is called the single-payment compound amount factor (SPCAF), but it is usually
referred to as the F P factor. This is the conversion factor that, when multiplied by P , yields
the future amount F of an initial amount P after n years at interest rate i . The cash flow diagram
is seen in Figure 2–1 a .
Reverse the situation to determine the P value for a stated amount F that occurs n periods
in the future. Simply solve Equation [2.2] for P .
P F [
1 ————
(1 i) n ]
F(1 i) n [2.3]
The expression (1 i ) n is known as the single-payment present worth factor (SPPWF), or the
P F factor. This expression determines the present worth P of a given future amount F after n
years at interest rate i . The cash flow diagram is shown in Figure 2–1 b .
Note that the two factors derived here are for single payments; that is, they are used to find the
present or future amount when only one payment or receipt is involved.
A standard notation has been adopted for all factors. The notation includes two cash flow symbols,
the interest rate, and the number of periods. It is always in the general form ( X Y , i , n ). The
letter X represents what is sought, while the letter Y represents what is given. For example, F P
means fi nd F when given P. The i is the interest rate in percent, and n represents the number of
periods involved.
Using this notation, ( F P ,6%,20) represents the factor that is used to calculate the future
amount F accumulated in 20 periods if the interest rate is 6% per period. The P is given. The
standard notation, simpler to use than formulas and factor names, will be used hereafter.
Table 2–1 summarizes the standard notation and equations for the F P and P F factors. This
information is also included inside the front cover.
TABLE 2–1
FP and PF Factors: Notation and Equations
Factor Standard Notation Equation Excel
Notation Name Find/Given Equation with Factor Formula Function
(FP,i,n)
(PF,i,n)
Single-payment
compound amount
Single-payment
present worth
FP F P(FP,i,n) F P(1 i) n FV(i%,n,,P)
PF P F(PF,i,n) P F (1 i) n PV(i%,n,,F)

2.1 Single-Amount Factors (FP and PF ) 41
To simplify routine engineering economy calculations, tables of factor values have been prepared
for interest rates from 0.25% to 50% and time periods from 1 to large n values, depending
on the i value. These tables, found at the rear of the book, have a colored edge for easy identification.
They are arranged with factors across the top and the number of periods n down the left side.
The word discrete in the title of each table emphasizes that these tables utilize the end-of-period
convention and that interest is compounded once each interest period. For a given factor, interest
rate, and time, the correct factor value is found at the intersection of the factor name and n . For
example, the value of the factor ( P F ,5%,10) is found in the P F column of Table 10 at period 10
as 0.6139. This value is determined by using Equation [2.3].
(PF,5%,10) ———— 1
(1 i) n
———— 1
(1.05) 10
1 ———
1.6289 0.6139
For spreadsheets, a future value F is calculated by the FV function using the format
FV(i%,n,,P) [2.4]
A present amount P is determined using the PV function with the format
PV(i%,n,,F) [2.5]
These functions are included in Table 2–1. Refer to Appendix A or Excel online help for more
information on the use of FV and PV functions.
EXAMPLE 2.1
Sandy, a manufacturing engineer, just received a year-end bonus of $10,000 that will be invested
immediately. With the expectation of earning at the rate of 8% per year, Sandy hopes to take the
entire amount out in exactly 20 years to pay for a family vacation when the oldest daughter is due
to graduate from college. Find the amount of funds that will be available in 20 years by using
(a) hand solution by applying the factor formula and tabulated value and (b) a spreadsheet function.
Solution
The cash flow diagram is the same as Figure 2–1a. The symbols and values are
P $10,000 F ? i 8% per year n 20 years
(a) Factor formula: Apply Equation [2.2] to find the future value F. Rounding to four decimals,
we have
F P(1 i) n 10,000(1.08) 20 10,000(4.6610)
$46,610
Standard notation and tabulated value: Notation for the FP factor is (FP,i%,n).
F P(FP,8%,20) 10,000(4.6610)
$46,610
Table 13 provides the tabulated value. Round-off errors can cause a slight difference in
the final answer between these two methods.
(b) Spreadsheet: Use the FV function to find the amount 20 years in the future. The format is
that shown in Equation [2.4]; the numerical entry is FV(8%,20,,10000). The spreadsheet
will appear similar to that in the right side of Figure 1–13, with the answer
($46,609.57) displayed. (You should try it on your own computer now.) The FV function
has performed the computation in part (a) and displayed the result.
The equivalency statement is: If Sandy invests $10,000 now and earns 8% per year every year
for 20 years, $46,610 will be available for the family vacation.

42 Chapter 2 Factors: How Time and Interest Affect Money
EXAMPLE 2.2 The Cement Factory Case
As discussed in the introduction to this chapter, the Houston American Cement factory will
require an investment of $200 million to construct. Delays beyond the anticipated implementation
year of 2012 will require additional money to construct the factory. Assuming that the cost
of money is 10% per year, compound interest, use both tabulated factor values and spreadsheet
functions to determine the following for the board of directors of the Brazilian company
that plans to develop the plant.
(a) The equivalent investment needed if the plant is built in 2015.
(b) The equivalent investment needed had the plant been constructed in the year 2008.
Solution
Figure 2–2 is a cash flow diagram showing the expected investment of $200 million ($200 M)
in 2012, which we will identify as time t 0. The required investments 3 years in the future
and 4 years in the past are indicated by F 3 ? and P 4 ?, respectively.
P −4 =?
F 3 =?
Figure 2–2
Cash flow diagram for
Example 2.2a and b.
PE
−4 −3 −2 −1 0 1 2 3 t
2008 2009 2010 2011 2012 2013 2014 2015 Year
$200 M
(a) To find the equivalent investment required in 3 years, apply the FP factor. Use $1 million
units and the tabulated value for 10% interest (Table 15).
F 3 P(FP,i,n) 200(FP,10%,3) 200(1.3310)
$266.2 ($266,200,000)
Now, use the FV function on a spreadsheet to find the same answer, F 3 $266.20 million.
(Refer to Figure 2–3, left side.)
FV(10%,3,,200)
PV(10%,4,,200)
Figure 2–3
Spreadsheet functions for Example 2.2.
(b) The year 2008 is 4 years prior to the planned construction date of 2012. To determine the
equivalent cost 4 years earlier, consider the $200 M in 2012 (t 0) as the future value F
and apply the PF factor for n 4 to find P 4 . (Refer to Figure 2–2.) Table 15 supplies
the tabulated value.
P 4 F(PF,i,n) 200(PF,10%,4) 200(0.6830)
$136.6 ($136,600,000)
The PV function PV(10%,4,,200) will display the same amount as shown in Figure
2–3, right side.
This equivalence analysis indicates that at $136.6 M in 2008, the plant would have cost about
68% as much as in 2012, and that waiting until 2015 will cause the price tag to increase about
33% to $266 M.

2.2 Uniform Series Present Worth Factor and Capital Recovery Factor (PA and AP) 43
2.2 Uniform Series Present Worth Factor and
Capital Recovery Factor ( P A and A P )
The equivalent present worth P of a uniform series A of end-of-period cash flows (investments)
is shown in Figure 2–4 a . An expression for the present worth can be determined by considering
each A value as a future worth F , calculating its present worth with the P F factor, Equation [2.3],
and summing the results.
P A [
1 ————
(1 i) ] A 1
[ 1 ————
(1 i) ] A 1
[ 2
(1 i) ] A 1
[ n1 ————
(1 i) ] n
A [
1 ————
————
(1 i) ] . . .
3
The terms in brackets are the PF factors for years 1 through n , respectively. Factor out A .
P A [
———— 1
(1 i) 1
1 ————
(1 i) 1
2 ————
(1 i) . . .
1
3 ————
(1 i) 1
n1 ————
(1 i) ] n [2.6]
To simplify Equation [2.6] and obtain the PA factor, multiply the n -term geometric progression
in brackets by the ( P F , i %,1) factor, which is 1(1 i ). This results in Equation [2.7]. Now
subtract the two equations, [2.6] from [2.7], and simplify to obtain the expression for P when
i 0 (Equation [2.8]).
——— P
1 i A ———— 1
[(1 i) 1
2 ————
(1 i) 1
3 ————
(1 i) . . . 1
4 ————
(1 i) n
——— 1
1 i P A ———— 1
[(1 i) 1
2 ————
(1 i) . . . 1
3 ————
(1 i) n ———— 1
(1 i) ] n1
P A [
———— 1
(1 i) 1
1 ————
(1 i) . . .
1
2 ————
(1 i) 1
n1 ————
(1 i) ]
n
——— i
1 i P A ———— 1
[(1 i) 1
n1 ————
(1 i) ] 1
P ——
i A ———— 1
[(1 i) n 1 ]
———— 1
(1 i ) ] [2.7]
n1
P A [ (1 i)n 1
——————
i(1 i) n ] i 0 [2.8]
The term in brackets in Equation [2.8] is the conversion factor referred to as the uniform series
present worth factor (USPWF). It is the PA factor used to calculate the equivalent P value in
year 0 for a uniform end-of-period series of A values beginning at the end of period 1 and extending
for n periods. The cash flow diagram is Figure 2–4 a .
P =?
i = given
P = given
i = given
0 1 2 n – 2 n – 1 n
0
1 2
n – 2 n – 1 n
A = given
(a)
Figure 2–4
Cash flow diagrams used to determine (a) P, given a uniform series A, and (b) A, given a present worth P.
A =?
(b)

44 Chapter 2 Factors: How Time and Interest Affect Money
TABLE 2–2
PA and AP Factors: Notation and Equations
Factor
Notation Name Find/Given
Factor
Formula
(PA,i,n) Uniform series PA (1 i) n 1
—————
present worth
i(1 i) n
(AP,i,n) Capital recovery AP i(1 i ) n
—————
(1 i ) n − 1
Standard
Notation Equation
P A(PA,i,n)
A P(AP,i,n)
Excel
Function
PV(i%,n,A)
PMT(i%,n,P)
To reverse the situation, the present worth P is known and the equivalent uniform series
amount A is sought (Figure 2–4 b ). The first A value occurs at the end of period 1, that is, one
period after P occurs. Solve Equation [2.8] for A to obtain
A P [
i(1 i ) n
——————
(1 i ) n 1 ] [2.9]
The term in brackets is called the capital recovery factor (CRF), or AP factor. It calculates the
equivalent uniform annual worth A over n years for a given P in year 0, when the interest
rate is i.
Placement of P
The PA and AP factors are derived with the present worth P and the first uniform annual
amount A one year (period) apart. That is, the present worth P must always be located one
period prior to the first A .
The factors and their use to find P and A are summarized in Table 2–2 and inside the front cover.
The standard notations for these two factors are ( P A , i %, n ) and ( A P , i %, n ). Tables at the end of
the text include the factor values. As an example, if i 15% and n 25 years, the P A factor
value from Table 19 is ( P A ,15%,25) 6.4641. This will find the equivalent present worth at
15% per year for any amount A that occurs uniformly from years 1 through 25.
Spreadsheet functions can determine both P and A values in lieu of applying the PA and A P
factors. The PV function calculates the P value for a given A over n years and a separate F value
in year n , if it is given. The format, is
PV ( i %, n , A , F ) [2.10]
Similarly, the A value is determined by using the PMT function for a given P value in year 0 and
a separate F , if given. The format is
Table 2–2 includes the PV and PMT functions.
PMT ( i %,n , P , F ) [2.11]
EXAMPLE 2.3
How much money should you be willing to pay now for a guaranteed $600 per year for 9 years
starting next year, at a rate of return of 16% per year?
Solution
The cash flows follow the pattern of Figure 2–4 a , with A $600, i 16%, and n 9. The
present worth is
P 600( P A ,16%,9) 600(4.6065) $2763.90
The PV function PV(16%,9,600) entered into a single spreadsheet cell will display the
answer P ($2763.93).

2.2 Uniform Series Present Worth Factor and Capital Recovery Factor (PA and AP) 45
EXAMPLE 2.4 The Cement Factory Case
As mentioned in the chapter introduction of this case, the Houston American Cement plant
may generate a revenue base of $50 million per year. The president of the Brazilian parent
company Votorantim Cimentos may have reason to be quite pleased with this projection for
the simple reason that over the 5-year planning horizon, the expected revenue would total
$250 million, which is $50 million more than the initial investment. With money worth
10% per year, address the following question from the president: Will the initial investment
be recovered over the 5-year horizon with the time value of money considered? If so, by how
much extra in present worth funds? If not, what is the equivalent annual revenue base required
for the recovery plus the 10% return on money? Use both tabulated factor values and spreadsheet
functions.
PE
Solution
Tabulated value: Use the P A factor to determine whether A $50 million per year for
n 5 years starting 1 year after the plant’s completion ( t 0) at i 10% per year is equivalently
less or greater than $200 M. The cash flow diagram is similar to Figure 2–4 a , where the
first A value occurs 1 year after P . Using $1 million units and Table 15 values,
P 50( P A ,10%,5) 50(3.7908)
$189.54 ($189,540,000)
The present worth value is less than the investment plus a 10% per year return, so the president
should not be satisfied with the projected annual revenue.
To determine the minimum required to realize a 10% per year return, use the A P factor.
The cash flow diagram is the same as Figure 2–4 b , where A starts 1 year after P at t 0 and
n 5.
A 200( A P ,10%,5) 200(0.26380)
$52.76 per year
The plant needs to generate $52,760,000 per year to realize a 10% per year return over
5 years.
Spreadsheet: Apply the PV and PMT functions to answer the question. Figure 2–5 shows
the use of PV( i %, n , A , F ) on the left side to find the present worth and the use of
PMT( i %, n , P , F ) on the right side to determine the minimum A of $52,760,000 per year.
Because there are no F values, it is omitted from the functions. The minus sign placed before
each function name forces the answer to be positive, since these two functions always display
the answer with the opposite sign entered on the estimated cash flows.
PV(10%,5,50)
PMT(10%,5,200)
Figure 2–5
Spreadsheet functions to find P and A for the cement factory case, Example 2.4.

46 Chapter 2 Factors: How Time and Interest Affect Money
2.3 Sinking Fund Factor and Uniform Series Compound
Amount Factor ( A F and F A )
The simplest way to derive the A F factor is to substitute into factors already developed. If P
from Equation [2.3] is substituted into Equation [2.9], the following formula results.
A F [
1 ————
—————
(1 i ) n − 1 ]
(1 i )
n ] [
i (1 i ) n
A F [
————— i
(1 i ) n 1 ] [2.12]
The expression in brackets in Equation [2.12] is the A F or sinking fund factor. It determines
the uniform annual series A that is equivalent to a given future amount F . This is shown graphically
in Figure 2–6 a , where A is a uniform annual investment.
Placement of F
The uniform series A begins at the end of year (period) 1 and continues through the year of
the given F. The last A value and F occur at the same time.
Equation [2.12] can be rearranged to find F for a stated A series in periods 1 through n (Figure
2–6 b ).
F A [ (1 i ) n 1
——————
i ] [2.13]
The term in brackets is called the uniform series compound amount factor (USCAF), or F A factor.
When multiplied by the given uniform annual amount A , it yields the future worth of the uniform
series. It is important to remember that the future amount F occurs in the same period as the last A .
Standard notation follows the same form as that of other factors. They are ( F A , i , n ) and
( A F , i , n ). Table 2–3 summarizes the notations and equations, as does the inside front cover.
As a matter of interest, the uniform series factors can be symbolically determined by using an
abbreviated factor form. For example, F A ( F P )( P A ), where cancellation of the P is correct.
Using the factor formulas, we have
( F A , i , n ) [(1 i ) n ] [ (1 i ) n 1
——————
i (1 i ) ] n (1 i ) n 1
——————
i
For solution by spreadsheet, the FV function calculates F for a stated A series over n years.
The format is
FV ( i %, n , A , P ) [2.14]
The P may be omitted when no separate present worth value is given. The PMT function determines
the A value for n years, given F in year n and possibly a separate P value in year 0. The
format is
PMT ( i %, n , P , F ) [2.15]
If P is omitted, the comma must be entered so the function knows the last entry is an F value.
F = given
i = given
0 1 2 n – 2 n – 1 n
i = given
F =?
0 1 2 n – 2 n – 1 n
A =?
(a)
Figure 2–6
Cash flow diagrams to ( a ) find A, given F , and ( b ) find F, given A .
A = given
(b)

2.3 Sinking Fund Factor and Uniform Series Compound Amount Factor (AF and FA) 47
TABLE 2–3
FA and AF Factors: Notation and Equations
Factor
Notation Name Find/Given
Factor
Formula
(FA,i,n) Uniform series FA (1 i) n 1
—————
compound amount
i
(AF,i,n) Sinking fund AF ————— i
(1 i ) n − 1
Standard Notation
Equation
F A(FA,i,n)
A F(AF,i,n)
Excel
Functions
FV(i%,n,A)
PMT(i%,n,F)
EXAMPLE 2.5
The president of Ford Motor Company wants to know the equivalent future worth of a $1 million
capital investment each year for 8 years, starting 1 year from now. Ford capital earns at a
rate of 14% per year.
Solution
The cash flow diagram (Figure 2–7) shows the annual investments starting at the end of year 1
and ending in the year the future worth is desired. In $1000 units, the F value in year 8 is found
by using the FA factor.
F 1000( F A, 14%,8) 1000(13.2328) $13,232.80
i = 14%
F =?
0 1 2 3 4 5 6 7 8
A = $1000
Figure 2–7
Diagram to find F for a uniform series, Example 2.5.
EXAMPLE 2.6 The Cement Factory Case
Once again, consider the HAC case presented at the outset of this chapter, in which a projected
$200 million investment can generate $50 million per year in revenue for 5 years starting
1 year after start-up. A 10% per year time value of money has been used previously to determine
P , F , and A values. Now the president would like the answers to a couple of new questions
about the estimated annual revenues. Use tabulated values, factor formulas, or spreadsheet
functions to provide the answers.
( a ) What is the equivalent future worth of the estimated revenues after 5 years at 10% per year?
( b ) Assume that, due to the economic downturn, the president predicts that the corporation
will earn only 4.5% per year on its money, not the previously anticipated 10% per year.
What is the required amount of the annual revenue series over the 5-year period to be economically
equivalent to the amount calculated in ( a )?
PE
Solution
( a ) Figure 2–6 b is the cash flow diagram with A $50 million. Note that the last A value and
F ? both occur at the end of year n 5. We use tabulated values and the spreadsheet
function to find F in year 5.
Tabulated value: Use the FA factor and 10% interest factor table. In $1 million units, the
future worth of the revenue series is
F 50( FA ,10%,5) 50(6.1051)
$305.255 ($305,255,000)

48 Chapter 2 Factors: How Time and Interest Affect Money
FV(10%,5,50)
PMT(4.5%,5,,B5)
Figure 2–8
Spreadsheet functions to find F and A at i 4.5% for the cement factory case,
Example 2.6.
If the rate of return on the annual revenues were 0%, the total amount after 5 years would
be $250,000,000. The 10% per year return is projected to grow this value by 22%.
Spreadsheet: Apply the FV factor in the format FV(10%,5,50) to determine F
$305.255 million. Because there is no present amount in this computation, P is omitted
from the factor. See Figure 2–8, left side. (As before, the minus sign forces the FV function
to result in a positive value.)
( b ) The president of the Brazilian company planning to develop the cement plant in Georgia
is getting worried about the international economy. He wants the revenue stream to generate
the equivalent that it would at a 10% per year return, that is, $305.255 million, but
thinks that only a 4.5% per year return is achievable.
Factor formula: The AF factor will determine the required A for 5 years. Since the factor
tables do not include 4.5%, use the formula to answer the question. In $1 million units,
A 305.255( AF ,4.5%,5) 305.255 [
0.045 ——————
(1.045) 5 1 ] 305.255(0.18279)
$55.798
The annual revenue requirement grows from $50 million to nearly $55,800,000. This is a
significant increase of 11.6% each year.
Spreadsheet: It is easy to answer this question by using the PMT( i %, n ,,F) function with
i 4.5% and F $305.255 found in part ( a ). We can use the cell reference method
(described in Appendix A) for the future amount F . Figure 2–8, right side, displays the required
A of $55.798 per year (in $1 million units).
2.4 Factor Values for Untabulated i or n Values
Often it is necessary to know the correct numerical value of a factor with an i or n value that is
not listed in the compound interest tables in the rear of the book. Given specific values of i and
n , there are several ways to obtain any factor value.
• Use the formula listed in this chapter or the front cover of the book,
• Use an Excel function with the corresponding P , F , or A value set to 1.
• Use linear interpolation in the interest tables.
When the formula is applied, the factor value is accurate since the specific i and n values are
input. However, it is possible to make mistakes since the formulas are similar to each other, especially
when uniform series are involved. Additionally, the formulas become more complex
when gradients are introduced, as you will see in the following sections.
A spreadsheet function determines the factor value if the corresponding P , A , or F argument
in the function is set to 1 and the other parameters are omitted or set to zero. For example,
the P F factor is determined using the PV function with A omitted (or set to 0) and
F 1, that is, PV( i %,n,,1) or PV( i %, n ,0,1). A minus sign preceding the function identifier
causes the factor to have a positive value. Functions to determine the six common factors are
as follows.

2.4 Factor Values for Untabulated i or n Values 49
Factor To Do This Excel Function
PF Find P, given F. PV(i%,n,,1)
FP Find F, given P. FV(i%,n,,1)
PA Find P, given A. PV(i%,n,1)
AP Find A, given P. PMT(i%,n,1)
FA Find F, given A. FV(i%,n,1)
AF Find A, given F. PMT(i%,n,,1)
Figure 2–9 shows a spreadsheet developed explicitly to determine these factor values. When it is
made live in Excel, entering any combination of i and n displays the exact value for all six factors. The
values for i 3.25% and n 25 years are shown here. As we already know, these same functions will
determine a final P , A , or F value when actual or estimated cash flow amounts are entered.
Linear interpolation for an untabulated interest rate i or number of years n takes more time
to complete than using the formula or spreadsheet function. Also interpolation introduces some
level of inaccuracy, depending upon the distance between the two boundary values selected for
i or n , as the formulas themselves are nonlinear functions. Interpolation is included here for individuals
who wish to utilize it in solving problems. Refer to Figure 2–10 for a graphical description
of the following explanation. First, select two tabulated values ( x 1 and x 2 ) of the parameter
for which the factor is requested, that is, i or n , ensuring that the two values surround and are not
too distant from the required value x . Second, find the corresponding tabulated factor values
( f 1 and f 2 ). Third, solve for the unknown, linearly interpolated value f using the formulas below,
where the differences in parentheses are indicated in Figure 2–10 as a through c .
Enter requested i and n
Figure 2–9
Use of Excel functions to display
factor values for any i and n
values.
Factor value
axis
f 2
Table
Figure 2–10
Linear interpolation in factor
value tables.
c
f 1
f
Unknown
d
Table
Linear
assumption
a
Known
x 1
Required
x
Known
x 2
i or n
axis
b

50 Chapter 2 Factors: How Time and Interest Affect Money
f f 1 (x – x 1 ) ————
(x 2 – x 1 ) (f 2 – f 1 ) [2.16]
f f 1 a —
b c f 1 d [2.17]
The value of d will be positive or negative if the factor is increasing or decreasing, respectively,
in value between x 1 and x 2 .
EXAMPLE 2.7
Determine the PA factor value for i 7.75% and n 10 years, using the three methods described
previously.
Solution
Factor formula: Apply the formula from inside the front cover of the book for the PA factor.
Showing 5-decimal accuracy,
( P A ,7.75%,10) (1 i ) n 1
—————
i (1 i ) n (1.0775) 10 1
———————
0.0775(1.0775) 1.10947
10 ————
0.16348
6.78641
Spreadsheet: Utilize the spreadsheet function in Figure 2–9, that is, PV(7.75%,10,1), to
display 6.78641.
Linear interpolation: Use Figure 2–10 as a reference for this solution. Apply the Equation
[2.16] and [2.17] sequence, where x is the interest rate i , the bounding interest rates are
i 1 7% and i 2 8%, and the corresponding PA factor values are f 1 ( PA ,7%,10) 7.0236
and f 2 ( PA ,8%,10) 6.7101. With 4-place accuracy,
f f 1 (i i 1
———
)
( i 2 – i 1 ) ( f 2 – f 1 ) 7.0236 —————
(7.75 7)
(8 7)
7.0236 (0.75)(−0.3135) 7.0236 − 0.2351
6.7885
(6.7101 7.0236)
Comment
Note that since the P A factor value decreases as i increases, the linear adjustment is negative
at 0.2351. As is apparent, linear interpolation provides an approximation to the correct factor
value for 7.75% and 10 years, plus it takes more calculations than using the formula or spreadsheet
function. It is possible to perform two-way linear interpolation for untabulated i and n
values; however, the use of a spreadsheet or factor formula is recommended.
2.5 Arithmetic Gradient Factors (PG and AG)
Assume a manufacturing engineer predicts that the cost of maintaining a robot will increase by
$5000 per year until the machine is retired. The cash flow series of maintenance costs involves a
constant gradient, which is $5000 per year.
An arithmetic gradient series is a cash flow series that either increases or decreases by a constant
amount each period. The amount of change is called the gradient.
Formulas previously developed for an A series have year-end amounts of equal value. In the
case of a gradient, each year-end cash flow is different, so new formulas must be derived. First,
assume that the cash flow at the end of year 1 is the base amount of the cash flow series and,
therefore, not part of the gradient series. This is convenient because in actual applications, the
base amount is usually significantly different in size compared to the gradient. For example, if
you purchase a used car with a 1-year warranty, you might expect to pay the gasoline and insurance
costs during the first year of operation. Assume these cost $2500; that is, $2500 is the base
amount. After the first year, you absorb the cost of repairs, which can be expected to increase

2.5 Arithmetic Gradient Factors (PG and AG) 51
0 1 2 3 4
n –1 n
Time
Figure 2–11
Cash flow diagram of an
arithmetic gradient series.
$2500
$2700
$2900
$3100
$2500
+ (n – 2)200 $2500
+ (n – 1)200
0 1 2 3 4 5
n –1
G
2G
3G
4G
n
Time
Figure 2–12
Conventional arithmetic
gradient series without
the base amount.
(n –2)G
(n –1)G
each year. If you estimate that total costs will increase by $200 each year, the amount the second
year is $2700, the third $2900, and so on to year n , when the total cost is 2500 ( n 1)200. The
cash flow diagram is shown in Figure 2–11. Note that the gradient ($200) is first observed between
year 1 and year 2, and the base amount ($2500 in year 1) is not equal to the gradient.
Define the symbols G for gradient and CF n for cash flow in year n as follows.
G constant arithmetic change in cash flows from one time period to the next; G may be positive
or negative.
CF n base amount ( n 1) G [2.18]
It is important to realize that the base amount defines a uniform cash flow series of the size A that
occurs eash time period. We will use this fact when calculating equivalent amounts that involve
arithmetic gradients. If the base amount is ignored, a generalized arithmetic (increasing) gradient
cash flow diagram is as shown in Figure 2–12. Note that the gradient begins between years
1 and 2. This is called a conventional gradient .
EXAMPLE 2.8
A local university has initiated a logo-licensing program with the clothier Holister, Inc. Estimated
fees (revenues) are $80,000 for the first year with uniform increases to a total of $200,000
by the end of year 9. Determine the gradient and construct a cash flow diagram that identifies
the base amount and the gradient series.
Solution
The year 1 base amount is CF 1 $80,000, and the total increase over 9 years is
CF 9 CF 1 200,000 – 80,000 $120,000
Equation [2.18], solved for G , determines the arithmetic gradient.
G (CF 9 CF 1
——————
) ————
120,000
n 1 9 1
$15,000 per year

52 Chapter 2 Factors: How Time and Interest Affect Money
CF 9 =
G = $15,000
$200,000
$185,000
$170,000
$155,000
$140,000
$125,000
$110,000
CF 1 = $95,000
$80,000
0 1 2 3 4 5 6 7 8 9
Figure 2–13
Diagram for gradient series, Example 2.8.
Year
The cash flow diagram (Figure 2–13) shows the base amount of $80,000 in years 1 through 9
and the $15,000 gradient starting in year 2 and continuing through year 9.
The total present worth P T for a series that includes a base amount A and conventional arithmetic
gradient must consider the present worth of both the uniform series defined by A and the
arithmetic gradient series. The addition of the two results in P T .
P T P A P G [2.19]
where P A is the present worth of the uniform series only, P G is the present worth of the gradient
series only, and the or sign is used for an increasing ( G ) or decreasing ( G ) gradient,
respectively.
The corresponding equivalent annual worth A T is the sum of the base amount series annual
worth A A and gradient series annual worth A G , that is,
A T A A A G [2.20]
Three factors are derived for arithmetic gradients: the P G factor for present worth, the A G
factor for annual series, and the F G factor for future worth. There are several ways to derive
them. We use the single-payment present worth factor ( P F , i , n ), but the same result can be obtained
by using the F P , F A , or P A factor.
In Figure 2–12, the present worth at year 0 of only the gradient is equal to the sum of the present
worths of the individual cash flows, where each value is considered a future amount.
Factor out G and use the P F formula.
P G(PF,i,2) 2G(PF,i,3) 3G(PF,i,4) . . .
[(n 2)G](PF,i,n 1) [(n 1)G](PF,i,n)
P G [
———— 1
(1 i) 2
2 ————
(1 i) 3
3 ————
(1 i) . . .
4
Multiplying both sides of Equation [2.21] by (1 i ) 1 yields
———— n 2
(1 i) n 1
n1 ————
(1 i) ] n [2.21]
P (1 i) 1 G [
———— 1
(1 i) 2
1 ————
(1 i) 3
2 ————
(1 i) . . .
n 2
3 ————
(1 i) n 1
n2 ————
(1 i) ] [2.22]
n1
Subtract Equation [2.21] from Equation [2.22] and simplify.
iP G [
———— 1
(1 i) 1
1 ————
(1 i) . . .
2
———— 1
(1 i) 1
n1 ————
(1 i) n ]
G
[
n
————
(1 i) ] n [2.23]
The left bracketed expression is the same as that contained in Equation [2.6], where the
P A factor was derived. Substitute the closed-end form of the P A factor from Equation [2.8]

2.5 Arithmetic Gradient Factors (PG and AG) 53
i = given
P G =?
0 1 2 3 4 n –1
n
0 1 2 3 4
n –1
n
G
2G
3G
(a)
(n –2)G
(n –1)G
(b)
Figure 2–14
Conversion diagram from an arithmetic gradient to a present worth.
into Equation [2.23] and simplify to solve for P G , the present worth of the gradient series
only.
P G — G i [ (1 i)n 1
——————
i(1 i) n ———— n
(1 i) ] n [2.24]
Equation [2.24] is the general relation to convert an arithmetic gradient G (not including the
base amount) for n years into a present worth at year 0 . Figure 2–14 a is converted into the
equivalent cash flow in Figure 2–14 b . The arithmetic gradient present worth factor, or P G
factor, may be expressed in two forms:
(PG,i,n) — 1 i [ (1 + i)n 1
—————
i(1 i) n ———— n
(1 i) ] n
or (PG,i,n) (1 i)n in 1
————————
i 2 (1 i) n [2.25]
Remember: The conventional arithmetic gradient starts in year 2, and P is located in year 0.
Equation [2.24] expressed as an engineering economy relation is
P G G(PG,i,n) [2.26]
which is the rightmost term in Equation [2.19] to calculate total present worth. The G carries a
minus sign for decreasing gradients.
The equivalent uniform annual series A G for an arithmetic gradient G is found by multiplying
the present worth in Equation [2.26] by the ( A P , i , n ) formula. In standard notation form, the
equivalent of algebraic cancellation of P can be used.
In equation form,
A G G(PG,i,n)(AP,i,n)
G(AG,i,n)
A G — G i [ (1 i)n 1
—————
i(1 i) n ———— n
——————
(1 i) n 1 ]
(1 i) n ] [
i(1 i) n
Placement of
gradient P G
A G G [
— 1 i —————— n
(1 i) n 1 ] [2.27]
which is the rightmost term in Equation [2.20]. The expression in brackets in Equation [2.27] is
called the arithmetic gradient uniform series factor and is identified by ( A G , i , n ). This factor
converts Figure 2–15 a into Figure 2–15 b .
The P G and A G factors and relations are summarized inside the front cover. Factor values
are tabulated in the two rightmost columns of factor values at the rear of this text.

54 Chapter 2 Factors: How Time and Interest Affect Money
i = given
0 1 2 3 4 n –1
n
A G = ?
0 1 2 3 4
n –1
n
G
2G
(a)
3G
(n –2)G
(n –1)G
(b)
Figure 2–15
Conversion diagram of an arithmetic gradient series to an equivalent uniform annual series.
There is no direct, single-cell spreadsheet function to calculate P G or A G for an arithmetic
gradient. Use the NPV function to display P G and the PMT function to display A G after entering
all cash flows (base and gradient amounts) into contiguous cells. General formats for these functions
are
NPV(i%, second_cell:last_cell) first_cell [2.28]
PMT(i%, n, cell_with_P G ) [2.29]
The word entries in italic are cell references, not the actual numerical values. (See Appendix A,
Section A.2, for a description of cell reference formatting.) These functions are demonstrated in
Example 2.10.
An FG factor ( arithmetic gradient future worth factor ) to calculate the future worth F G of a
gradient series can be derived by multiplying the P G and F P factors. The resulting factor,
( F G , i , n ), in brackets, and engineering economy relation is
F G G [ ( — 1 i ) ( (1 i)n – 1
—————
i ) n ]
EXAMPLE 2.9
Neighboring parishes in Louisiana have agreed to pool road tax resources already designated
for bridge refurbishment. At a recent meeting, the engineers estimated that a total of
$500,000 will be deposited at the end of next year into an account for the repair of old and
safety-questionable bridges throughout the area. Further, they estimate that the deposits
will increase by $100,000 per year for only 9 years thereafter, then cease. Determine the
equivalent (a) present worth and (b) annual series amounts, if public funds earn at a rate
of 5% per year.
Solution
(a) The cash flow diagram of this conventional arithmetic gradient series from the perspective
of the parishes is shown in Figure 2–16. According to Equation [2.19], two computations
must be made and added: the first for the present worth of the base amount P A
and the second for the present worth of the gradient P G . The total present worth P T
occurs in year 0. This is illustrated by the partitioned cash flow diagram in Figure 2–17.
In $1000 units, the total present worth is
P T 500(PA,5%,10) 100(PG,5%,10)
500(7.7217) 100(31.6520)
$7026.05 ($7,026,050)

2.5 Arithmetic Gradient Factors (PG and AG) 55
0 1
2
3
4
5
6
7
8
9
10
$500
$600
$700
$800
$900
$1000
$1100
$1200
$1300
$1400
Figure 2–16
Cash flow series with a conventional arithmetic gradient (in $1000 units),
Example 2.9.
P A =?
A = $500
1 2 9 10
+
P G =?
G = $100
1 2 9 10
$100
Base
Gradient
$900
P T =?
1
2
3
4
P T = P A + P G
6
5
7
8
9
10
$500
$600
$700
$800
$900
Figure 2–17
Partitioned cash flow diagram (in $1000 units), Example 2.9.
$1000
$1100
$1200
$1300
$1400
(b) Here, too, it is necessary to consider the gradient and the base amount separately. The
total annual series A T is found by Equation [2.20] and occurs in years 1 through 10.
A T 500 100(AG,5%,10) 500 100(4.0991)
$909.91 per year ($909,910)
Comment
Remember: The PG and AG factors determine the present worth and annual series of the
gradient only. Any other cash flows must be considered separately.
If the present worth is already calculated [as in part (a)], P T can be multiplied by an AP
factor to get A T . In this case, considering round-off error,
A T P T (AP,5%,10) 7026.05(0.12950)
$909.873 ($909,873)

56 Chapter 2 Factors: How Time and Interest Affect Money
EXAMPLE 2.10 The Cement Factory Case
The announcement of the HAC cement factory states that the $200 million (M) investment is
planned for 2012. Most large investment commitments are actually spread out over several
years as the plant is constructed and production is initiated. Further investigation may determine,
for example, that the $200 M is a present worth in the year 2012 of anticipated investments
during the next 4 years (2013 through 2016). Assume the amount planned for 2013 is
$100 M with constant decreases of $25 M each year thereafter. As before, assume the time
value of money for investment capital is 10% per year to answer the following questions using
tabulated factors and spreadsheet functions, as requested below.
(a) In equivalent present worth values, does the planned decreasing investment series equal
the announced $200 M in 2012? Use both tabulated factors and spreadsheet functions.
(b) Given the planned investment series, what is the equivalent annual amount that will be
invested from 2013 to 2016? Use both tabulated factors and spreadsheet functions.
(c) (This optional question introduces Excel’s Goal Seek tool.) What must be the amount of
yearly constant decrease through 2016 to have a present worth of exactly $200 M in
2012, provided $100 M is expended in 2013? Use a spreadsheet.
Solution
(a) The investment series is a decreasing arithmetic gradient with a base amount of $100 M
in year 1 (2013) and G $25 M through year 4 (2016). Figure 2–18 diagrams the cash
flows with the shaded area showing the constantly declining investment each year. The P T
value at time 0 at 10% per year is determined by using tables and a spreadsheet.
Tabulated factors: Equation [2.19] with the minus sign for negative gradients determines
the total present worth P T . Money is expressed in $1 million units.
P T P A P G 100(PA,10%,4) 25(PG,10%,4) [2.30]
100(3.1699) – 25(4.3781)
$207.537 ($207,537,000)
In present worth terms, the planned series will exceed the equivalent of $200 M in 2012
by approximately $7.5 M.
Spreadsheet: Since there is no spreadsheet function to directly display present worth for a
gradient series, enter the cash flows in a sequence of cells (rows or columns) and use the
NPV function to find present worth. Figure 2–19 shows the entries and function
NPV(i%,second_cell:last_cell). There is no first_cell entry here, because there is no
investment per se in year 0. The result displayed in cell C9, $207.534, is the total P T for
the planned series. (Note that the NPV function does not consider two separate series of
cash flows as is necessary when using tabulated factors.)
The interpretation is the same as in part (a); the planned investment series exceeds the
$200 M in present worth terms by approximately $7.5 M.
(b) Tabulated factors: There are two equally correct ways to find A T . First, apply Equation
[2.20] that utilizes the AG factor, and second, use the P T value obtained above and
the AP factor. Both relations are illustrated here, in $1 million units,
PE
P T ?
i = 10% per year
Figure 2–18
Cash flow diagram for decreasing
gradient in $1 million
units, Example 2.10.
2013
2014
2015
2016
Year
0
1
2
3
4
Time
Base
A $100
$25
$50
$100
Gradient
G $25

2.5 Arithmetic Gradient Factors (PG and AG) 57
Figure 2–19
Spreadsheet solution for
Example 2.10a and b.
Present worth of investments
NPV(10%,C5:C8)
Annual worth of investments
PMT(10%,4,C9)
Use Equation [2.20]:
Use P T :
A T 100 – 25(AG,10%,4) 100 25(1.3812)
$65.471 ($65,471,000 per year)
A T 207.537(AP,10%,4) 207.537(0.31547)
$65.471 per year
Spreadsheet: Apply the PMT function in Equation [2.29] to obtain the same A T
$65.471 per year (Figure 2–19).
(c) (Optional) The Goal Seek tool is described in Appendix A. It is an excellent tool to apply
when one cell entry must equal a specific value and only one other cell can change. This
is the case here; the NPV function (cell C9 in Figure 2–19) must equal $200, and the gradient
G (cell C1) is unknown. This is the same as stating P T 200 in Equation [2.30] and
solving for G. All other parameters retain their current value.
Figure 2–20 (top) pictures the same spreadsheet used previously with the Goal Seek template
added and loaded. When OK is clicked, the solution is displayed; G $26.721.
Refer to Figure 2–20 again. This means that if the investment is decreased by a constant
annual amount of $26.721 M, the equivalent total present worth invested over the 4 years
will be exactly $200 M.
Figure 2–20
Solution for arithmetic
gradient using Goal Seek,
Example 2.10c.
Present worth of investments:
NPV(10%,C5:C8)
Set up Goal Seek template
Solution for G = $26.721 to make
present worth exactly $200

58 Chapter 2 Factors: How Time and Interest Affect Money
2.6 Geometric Gradient Series Factors
It is common for annual revenues and annual costs such as maintenance, operations, and labor to
go up or down by a constant percentage, for example, 5% or 3% per year. This change occurs
every year on top of a starting amount in the first year of the project. A definition and description
of new terms follow.
A geometric gradient series is a cash flow series that either increases or decreases by a constant
percentage each period. The uniform change is called the rate of change.
g constant rate of change , in decimal form, by which cash flow values increase or decrease
from one period to the next. The gradient g can be or .
A 1 initial cash flow in year 1 of the geometric series
P g
present worth of the entire geometric gradient series, including the initial amount
A 1
Note that the initial cash flow A 1 is not considered separately when working with geometric
gradients.
Figure 2–21 shows increasing and decreasing geometric gradients starting at an amount A 1 in
time period 1 with present worth P g located at time 0. The relation to determine the total present
worth P g for the entire cash flow series may be derived by multiplying each cash flow in Figure
2–21 a by the P F factor 1(1 i )
n
.
A
P g ———— 1
(1 i) A 1 (1 g)
1 ————
(1 i) A 1 (1 g)2
2 —————
(1 i) . . . A 1 (1 g)n1
3 ——————
(1 i) n
A 1 ——— 1
[ 1 i 1 g (1 g)2
———— ————
2
(1 i) (1 i) . . .
(1 g)n1
3 —————
(1 i) ] n [2.31]
Multiply both sides by (1 g )(1 i ), subtract Equation [2.31] from the result, factor out P g ,
and obtain
P g ( ———
1 g
1 i 1 ) A (1 g)n
1 —————
[ (1 i) ——— 1
n1 1 i ]
Solve for P g and simplify.
P g A 1 [ 1 (———
1 g
1 i ) n
———————
i g ] g i [2.32]
The term in brackets in Equation [2.32] is the ( P A , g , i , n ) or geometric gradient series present
worth factor for values of g not equal to the interest rate i. When g i , substitute i for g in Equation
[2.31] and observe that the term 1/(1 + i) appears n times.
P g =?
i = given
g = given
P g =?
i = given
g = given
0 1 2
A 1
3
A 1 (1 + g)
A 1 (1 + g) 2
4
A 1 (1 + g) 3
n
0 1 2 3 4
A 1 (1 – g)
A n –1
1 (1 – g) 3
A 1 (1 – g) 2
A 1 (1 – g)
n
(a)
A 1 (1 + g) n –1
A 1
(b)
Figure 2–21
Cash flow diagram of (a) increasing and (b) decreasing geometric gradient series and present worth P g .

2.6 Geometric Gradient Series Factors 59
P g A 1 (
——— 1
(1 i ) ——— 1
(1 i ) ——— 1
(1 i ) ...
nA
P g ——— 1
(1 i)
——— 1
(1 i ) )
[2.33]
The (PA,g,i,n) factor calculates P g in period t 0 for a geometric gradient series starting in
period 1 in the amount A 1 and increasing by a constant rate of g each period.
The equation for P g and the ( P A , g , i , n ) factor formula are
Placement of
Gradient P g
(PA,g,i,n)
P g A 1 (PA,g,i,n) [2.34]
1 ( ———
1 g
1 i ) n
——————
i − g
n ———
1 i
g i
g i
[2.35]
It is possible to derive factors for the equivalent A and F values; however, it is easier to determine
the P g amount and then multiply by the A P or F P factor.
As with the arithmetic gradient series, there are no direct spreadsheet functions for geometric
gradient series. Once the cash flows are entered, P and A are determined using the NPV and PMT
functions, respectively.
EXAMPLE 2.11
A coal-fired power plant has upgraded an emission control valve. The modification costs only
$8000 and is expected to last 6 years with a $200 salvage value. The maintenance cost is expected
to be high at $1700 the first year, increasing by 11% per year thereafter. Determine the
equivalent present worth of the modification and maintenance cost by hand and by spreadsheet
at 8% per year.
Solution by Hand
The cash flow diagram (Figure 2–22) shows the salvage value as a positive cash flow and all
costs as negative. Use Equation [2.35] for g i to calculate P g . Total P T is the sum of three
present worth components.
P T =?
P g =?
$8000
P T 8000 P g 200( P F ,8%,6)
8000 1700 [ 1 (1.111.08) 6
———————
0.08 0.11 ] 200( P F ,8%,6)
8000 1700(5.9559) 126 $17,999
1
$1700
2
i =8%
g = 11%
3
4 5
6
$1700(1.11)
$1700(1.11) 2
$1700(1.11) 3
$1700(1.11) 4
Figure 2–22
Cash flow diagram of a geometric
gradient, Example 2.11.
$200
$1700(1.11) 5

60 Chapter 2 Factors: How Time and Interest Affect Money
Solution by Spreadsheet
Figure 2–23 details the spreadsheet operations to find the geometric gradient present worth P g
and total present worth P T . To obtain P T $17,999, three components are summed—first
cost, present worth of estimated salvage in year 6, and P g . Cell tags detail the relations for the
second and third components; the first cost occurs at time 0.
Comment
The relation that calculates the ( PA,g,i%,n ) factor is rather complex, as shown in the cell tag
and formula bar for C9. If this factor is used repeatedly, it is worthwhile using cell reference
formatting so that A 1 , i , g , and n values can be changed and the correct value is always obtained.
Try to write the relation for cell C9 in this format.
Present worth of salvage
PV(8%,6,,200)
Present worth of maintenance costs, Eq. [2.35]
1700 * ((1-((1.11)/(1.08))^6)/(0.08-0.11))
Figure 2–23
Geometric gradient and total present worth calculated via spreadsheet, Example 2.11.
EXAMPLE 2.12 The Cement Factory Case
Now let’s go back to the proposed Houston American Cement plant in Georgia. The revenue
series estimate of $50 million annually is quite optimistic, especially since there are many
other cement product plants operating in Florida and Georgia on the same limestone deposit.
(The website for the HAC plant shows where they are located currently; it is clear that keen
competition will be present.) Therefore, it is important to be sensitive in our analysis to possibly
declining and increasing revenue series, depending upon the longer-term success of the
plant’s marketing, quality, and reputation. Assume that revenue may start at $50 million by the
end of the first year, but then decreases geometrically by 12% per year through year 5. Determine
the present worth and future worth equivalents of all revenues during this 5-year time
frame at the same rate used previously, that is, 10% per year.
Solution
The cash flow diagram appears much like Figure 2–21 b , except that the arrows go up for revenues.
In year 1, A 1 $50 M and revenues decrease in year 5 to
A 1 (1 g ) n 1
50 M(1 0.12) 5–1 50 M(0.88) 4 $29.98 M
First, we determine P g in year 0 using Eq. [2.35] with i 0.10 and g 0.12, then we calculate
F in year 5. In $1 million units,
1
P g 50 [ (
—— 0.88
$152.80
1.10 ) 5
———————
0.10 (0.12) ] 50[3.0560]
F 152.80( FP ,10%,5) 152.80(1.6105)
$246.08
This means that the decreasing revenue stream has a 5-year future equivalent worth of
$246.080 M. If you look back to Example 2.6, we determined that the F in year 5 for the
PE

2.7 Determining i or n for Known Cash Flow Values 61
uniform revenue series of $50 M annually is $305.255 M. In conclusion, the 12% declining
geometric gradient has lowered the future worth of revenue by $59.175 M, which is a sizable
amount from the perspective of the owners of Votorantim Cimentos North America, Inc.
2.7 Determining i or n for Known Cash Flow Values
When all the cash flow values are known or have been estimated, the i value (interest rate or rate
of return) or n value (number of years) is often the unknown. An example for which i is sought
may be stated as follows: A company invested money to develop a new product. After the net
annual income series is known following several years on the market, determine the rate of return
i on the investment. There are several ways to find an unknown i or n value, depending upon the
nature of the cash flow series and the method chosen to find the unknown. The simplest case involves
only single amounts ( P and F ) and solution utilizing a spreadsheet function. The most
difficult and complex involves finding i or n for irregular cash flows mixed with uniform and
gradient series utilizing solution by hand and calculator. The solution approaches are summarized
below, followed by examples.
Single Amounts— P and F Only
Hand or Calculator Solution Set up the equivalence relation and (1) solve for the variable
using the factor formula, or (2) find the factor value and interpolate in the tables.
Spreadsheet Solution Use the IRR or RATE function to find i or the NPER function to find
n . (See below and Appendix A for details.)
Uniform Series— A Series
Hand or Calculator Solution Set up the equivalence relation using the appropriate factor
( P/A , A/P , F/A , or A/F ), and use the second method mentioned above.
Spreadsheet Solution Use the IRR or RATE function to find i or the NPER function to find n .
Mixed A Series, Gradients, and/or Isolated Values
Hand or Calculator Solution Set up the equivalence relation and use (1) trial and error or
(2) the calculator functions.
Spreadsheet Solution Use the IRR or RATE function to find i or the NPER function to find
n . (This is the recommended approach.)
Besides the PV, FV, and NPV functions, other spreadsheet functions useful in determining i
are IRR (internal rate of return) and RATE, and NPER (number of periods) to find n . The formats
are shown here and the inside front cover with a detailed explanation in Appendix A. In all three
of these functions, at least one cash flow entry must have a sign opposite that of others in order
to find a solution.
IRR(first_cell:last_cell) [2.36]
To use IRR to find i , enter all cash flows into contiguous cells, including zero values.
RATE( n , A , P , F ) [2.37]
The single-cell RATE function finds i when an A series and single P and/or F values are involved.
NPER( i %, A , P , F ) [2.38]
NPER is a single-cell function to find n for single P and F values, or with an A series.

62 Chapter 2 Factors: How Time and Interest Affect Money
EXAMPLE 2.13
If Laurel made a $30,000 investment in a friend’s business and received $50,000 5 years later,
determine the rate of return.
Solution
Since only single amounts are involved, i can be determined directly from the P / F factor.
P F ( P F , i , n ) F ———— 1
(1 i ) n
30,000 50,000 ———— 1
(1 i ) 5
0.600 ———— 1
(1 i ) 5
i ( 1 ——
0.6 ) 0.2 1 0.1076 (10.76%)
Alternatively, the interest rate can be found by setting up the standard P F relation, solving for
the factor value, and interpolating in the tables.
P F ( P F , i , n )
30,000 50,000( P F , i ,5)
( P F , i ,5) 0.60
From the interest tables, a P / F factor of 0.6000 for n 5 lies between 10% and 11%. Interpolate
between these two values to obtain i 10.76%.
EXAMPLE 2.14
Pyramid Energy requires that for each of its offshore wind power generators $5000 per year
be placed into a capital reserve fund to cover unexpected major rework on field equipment.
In one case, $5000 was deposited for 15 years and covered a rework costing $100,000 in
year 15. What rate of return did this practice provide to the company? Solve by hand and
spreadsheet.
Solution by Hand
The cash flow diagram is shown in Figure 2–24 . Either the A F or F A factor can be used.
Using A F ,
A F ( A F , i , n )
5000 100,000( A F , i ,15)
( A F , i ,15) 0.0500
From the AF interest tables for 15 years, the value 0.0500 lies between 3% and 4%. By interpolation,
i 3.98%.
i = ?
F = $100,000
Figure 2–24
Diagram to determine the rate
of return, Example 2.14.
0
1
2 3 4 5 6 7 8 9 10 11 12 13 14 15
A = $5000

2.7 Determining i or n for Known Cash Flow Values 63
Solution by Spreadsheet
Refer to the cash flow diagram ( Figure 2–24 ) while completing the spreadsheet ( Figure 2–25 ).
A single-cell solution using the RATE function can be applied since A $5000 occurs
each year and F $100,000 takes place in the last year of the series. The function
RATE(15,5000,,100000) displays the value i 3.98%. This function is fast, but it allows
only limited sensitivity analysis because all the A values have to change by the same amount.
The IRR function is much better for answering “what if ” questions.
To apply the IRR function, enter the value 0 in a cell (for year 0), followed by –5000 for
14 years and in year 15 enter 95,000 ( Figure 2–25 ). In any cell enter the IRR function. The
answer i 3.98% is displayed. It is advisable to enter the year numbers 0 through n (15 in
this example) in the column immediately to the left of the cash flow entries. The IRR function
does not need these numbers, but it makes the cash flow entry activity easier and more
accurate. Now any cash flow can be changed, and a new rate will be displayed immediately
via IRR.
i using RATE function
RATE(15,-5000,,100000)
Figure 2–25
Use of RATE and IRR
functions to determine
i value for a uniform
series, Example 2.14.
i using IRR function
IRR(E2:E17))
EXAMPLE 2.15 The Cement Factory Case
From the introductory comments about the HAC plant, the annual revenue is planned to be
$50 million. All analysis thus far has taken place at 10% per year; however, the parent company
has made it clear that its other international plants are able to show a 20% per year return
on the initial investment. Determine the number of years required to generate 10%, 15%, and
20% per year returns on the $200 million investment at the Georgia site.
PE
Solution
If hand solution is utilized, the present worth relation can be established and the n values
interpolated in the tables for each of the three rate of return values. In $1 million units, the
relation is
( PA , i %, n ) 4.00
P 200 50( PA , i %, n ) ( i 10%, 15%, 20%)
This is a good opportunity to utilize a spreadsheet and repeated NPER functions from
Equation [2.38], since several i values are involved. Figure 2–26 shows the single-cell
NPER( i %,50,200) function for each rate of return. The number of years (rounded up) to
produce at least the required returns are
Return, i% Years
10 6
15 7
20 9

64 Chapter 2 Factors: How Time and Interest Affect Money
Figure 2–26
Use of NPER function to find n values for various rate of return requirements,
Example 2.15.
CHAPTER SUMMARY
Formulas and factors derived and applied in this chapter perform equivalence calculations for
present, future, annual, and gradient cash flows. Capability in using these formulas and their
standard notation manually and with spreadsheets is critical to complete an engineering economy
study. Using these formulas and spreadsheet functions, you can convert single cash flows into
uniform cash flows, gradients into present worths, and much more. Additionally, you can solve
for rate of return i or time n .
PROBLEMS
Use of Interest Tables
2.1 Look up the numerical value for the following factors
from the interest tables.
1. ( PF ,6%,8)
2. ( AP ,10%,10)
3. ( AG ,15%,20)
4. ( AF ,2%,30)
5. ( PG ,35%,15)
Determination of F , P , and A
2.2 How much can Haydon Rheosystems, Inc., afford
to spend now on an energy management system if
the software will save the company $21,300 per
year for the next 5 years? Use an interest rate of
10% per year.
2.3 A manufacturer of off-road vehicles is considering
the purchase of dual-axis inclinometers for installation
in a new line of tractors. The distributor of the
inclinometers is temporarily overstocked and is offering
them at a 40% discount from the regular cost
of $142. If the purchaser gets them now instead of
2 years from now, which is when they will be
needed, what is the present worth of the savings per
unit? The company would pay the regular price, if
purchased in 2 years. Assume the interest rate is
10% per year.
2.4 The Moller Skycar M400 is a flying car known
as a personal air vehicle (PAV) that is expected
to be FAA-certified by December 31, 2011. The
cost is $985,000, and a $100,000 deposit will hold
one of the first 100 “cars.” Assume a buyer pays
the $885,000 balance 3 years after making the
$100,000 deposit. At an interest rate of 10% per
year, what is the effective total cost of the PAV in
year 3?
2.5 A family that won a $100,000 prize on America’s
Funniest Home Videos decided to put one-half of
the money in a college fund for their child who
was responsible for the prize. If the fund earned
interest at 6% per year, how much was in the account
14 years after it was started?
2.6 One of the biggest vulnerabilities in a control system
is network devices, such as Ethernet-based
network switches that are located in unsecured
locations and accessible to everyone. DeltaX
switches, manufactured by Dahne Security, allow
the user to automatically lock and unlock the port
access to all switches in the network. The company
is considering expanding its manufacturing
lines now or doing it in 3 years. If the cost now
would be $1.9 million, what equivalent amount
could the company afford to spend in 3 years? The
interest rate is 15% per year.
2.7 A company that sells high-purity laboratory chemicals
is considering investing in new equipment

Problems 65
that will reduce cardboard costs by better matching
the size of the products to be shipped to the
size of the shipping container. If the new equipment
will cost $220,000 to purchase and install,
how much must the company save each year for
3 years in order to justify the investment, if the
interest rate is 10% per year?
2.8 Red Valve Co. of Carnegie, Pennsylvania, makes a
control pinch valve that provides accurate, repeatable
control of abrasive and corrosive slurries, outlasting
gate, plug, ball, and even satellite coated
valves. How much can the company afford to
spend now on new equipment in lieu of spending
$75,000 four years from now? The company’s rate
of return is 12% per year.
2.9 If GHD Plastics purchases a new building now for
$1.3 million for its corporate headquarters, what
must the building be worth in 10 years? The company
expects all expenditures to earn a rate of return
of at least 18% per year.
2.10 CGK Rheosystems makes high-performance rotational
viscometers capable of steady shear and
yield stress testing in a rugged, compact footprint.
How much could the company afford to spend now
on new equipment in lieu of spending $200,000
one year from now and $300,000 three years from
now, if the company uses an interest rate of 15%
per year?
2.11 Five years ago a consulting engineer purchased a
building for company offices constructed of bricks
that were not properly fired. As a result, some of
the bricks were deteriorated from their exposure to
rain and snow. Because of the problem with the
bricks, the selling price of the building was 25%
below the price of comparable, structurally sound
buildings. The engineer repaired the damaged
bricks and arrested further deterioration by applying
an extra-strength solvent-based RTV elastomeric
sealant. This resulted in restoring the building
to its fair market value. If the depressed
purchase price of the building was $600,000 and
the cost of getting it repaired was $25,000, what is
the equivalent value of the “forced appreciation”
today, if the interest rate is 8% per year?
2.12 Metso Automation, which manufactures addressable
quarter-turn electric actuators, is planning to
set aside $100,000 now and $150,000 one year
from now for possible replacement of the heating
and cooling systems in three of its larger manufacturing
plants. If the replacement won’t be needed
for 4 years, how much will the company have in
the account, if it earns interest at a rate of 8% per
year?
2.13 Syringe pumps often fail because reagents adhere
to the ceramic piston and deteriorate the seal. Trident
Chemical developed an integrated polymer
dynamic seal that provides a higher sealing force
on the sealing lip, resulting in extended seal life.
One of Trident’s customers expects to reduce
downtime by 30% as a result of the new seal design.
If lost production would have cost the company
$110,000 per year for the next 4 years, how
much could the company afford to spend now on
the new seals, if it uses an interest rate of 12% per
year?
2.14 China spends an estimated $100,000 per year on
cloud seeding efforts, which includes using antiaircraft
guns and rocket launchers to fill the sky
with silver iodide. In the United States, utilities
that run hydroelectric dams are among the most
active cloud seeders, because they believe it is a
cost-effective way to increase limited water supplies
by 10% or more. If the yields of cash crops
will increase by 4% each year for the next 3 years
because of extra irrigation water captured behind
dams during cloud seeding, what is the maximum
amount the farmers should spend now on the cloud
seeding activity? The value of the cash crops without
the extra irrigation water would be $600,000
per year. Use an interest rate of 10% per year.
2.15 The Public Service Board (PSB) awarded two contracts
worth a combined $1.07 million to improve
(i.e., deepen) a retention basin and reconstruct the
spillway that was severely damaged in a flood
2 years ago. The PSB said that, because of the weak
economy, the bids came in $950,000 lower than engineers
expected. If the projects are assumed to
have a 20-year life, what is the annual worth of the
savings at an interest rate of 6% per year?
2.16 The National Highway Traffic Safety Administration
raised the average fuel efficiency standard to
35.5 miles per gallon for cars and light trucks by
the year 2016. The rules will cost consumers an
average of $434 extra per vehicle in the 2012
model year. If a person purchases a new car in
2012 and keeps it for 5 years, how much must be
saved in fuel costs each year to justify the extra
cost? Use an interest rate of 8% per year.
2.17 In an effort to reduce childhood obesity by reducing
the consumption of sugared beverages,
some states have imposed taxes on soda and
other soft drinks. A survey by Roland Sturm of
7300 fifth-graders revealed that if taxes averaged
4 cents on each dollar’s worth of soda, no real
difference in overall consumption was noticed.
However, if taxes were increased to 18 cents on
the dollar, Sturm calculated they would make a

66 Chapter 2 Factors: How Time and Interest Affect Money
significant difference. For a student who consumes
100 sodas per year, what is the future worth
of the extra cost from 4 cents to 18 cents per
soda? Assume the student consumes sodas from
grade 5 through graduation in grade 12. Use an
interest rate of 6% per year.
2.18 The Texas Tomorrow Fund (TTF) is a program
started in 1996 in Texas wherein parents could prepay
their child's college tuition when the child was
young. Actuaries set the price based on costs and
investment earnings at that time. Later, the Texas
legislature allowed universities to set their own tuition
rates; tuition costs jumped dramatically. The
cost for entering a newborn in 1996 was $10,500.
If the TTF fund grew at a rate of 4% per year,
while tuition costs increased at 7% per year, determine
the state’s shortfall when a newborn enters
college 18 years later.
2.19 Henry Mueller Supply Co. sells tamperproof,
normally open thermostats (i.e., thermostat closes
as temperature rises). Annual cash flows are
shown in the table below. Determine the future
worth of the net cash flows at an interest rate of
10% per year.
Year 1 2 3 4 5 6 7 8
Income, $1000 200 200 200 200 200 200 200 200
Cost, $1000 90 90 90 90 90 90 90 90
2.20 A company that makes self-clinching fasteners expects
to purchase new production-line equipment
in 3 years. If the new units will cost $350,000, how
much should the company set aside each year, if
the account earns 10% per year?
Factor Values
2.21 Find the numerical value of the following factors
using ( a ) interpolation and ( b ) the formula.
1. ( AP ,13%,15)
2. ( PG ,27%,10)
2.22 Find the numerical value of the following factors
using ( a ) interpolation, ( b ) the formula, and ( c ) a
spreadsheet function.
1. ( FP ,14%,62)
2. ( AF ,1%,45)
2.23 For the factor ( FP ,10%,43), find the percent difference
between the interpolated and formulacalculated
values, assuming the formula- calculated
value is the correct one.
2.24 For the factor ( FA ,15%,52), find the percent difference
between the interpolated and formula-
calculated values, assuming the formula- calculated
value is the correct one.
Arithmetic Gradient
2.25 Profits from recycling paper, cardboard, aluminum,
and glass at a liberal arts college have increased
at a constant rate of $1100 in each of the
last 3 years. If this year’s profit (end of year 1) is
expected to be $6000 and the profit trend continues
through year 5, ( a ) what will the profit be at
the end of year 5 and ( b ) what is the present
worth of the profit at an interest rate of 8% per
year?
2.26 A report by the Government Accountability Office
(GAO) shows that the GAO expects the
U.S. Postal Service to lose a record $7 billion at
the end of this year, and if the business model is
not changed, the losses will total $241 billion by
the end of year 10. If the losses increase uniformly
over the 10-year period, determine the
following:
(a) The expected increase in losses each year
(b)
(c)
The loss 5 years from now
The equivalent uniform worth of the losses at
an interest rate of 8% per year
2.27 Rolled ball screws are suitable for high-precision
applications such as water jet cutting. Their total
manufacturing cost is expected to decrease because
of increased productivity, as shown in the
table. Determine the equivalent annual cost at an
interest rate of 8% per year.
Year 1 2 3 4 5 6 7 8
Cost, $1000 200 195 190 185 180 175 170 165
2.28 Western Hydra Systems makes a panel milling machine
with a 2.7-m-diameter milling head that
emits low vibration and processes stress-relieved
aluminum panels measuring up to 6000 mm long.
The company wants to borrow money for a new
production/warehouse facility. If the company offers
to repay the loan with $60,000 in year 1 and
amounts increasing by $10,000 each year through
year 5, how much can the company borrow at an
interest rate of 10% per year?
2.29 GKX Industries expects sales of its hydraulic seals
(in inch and metric sizes) to increase according to
the cash flow sequence $70 4 k , where k is in
years and cash flow is in $1000.
(a) What is the amount of the cash flow in year 3?
(b) What is the future worth of the entire cash
flow series in year 10? Let i 10% per year.

Problems 67
2.30 For the cash flows below, determine the amount in year 1, if the annual worth in years 1 through 9 is $601.17 and
the interest rate is 10% per year.
Year 1 2 3 4 5 6 7 8 9
Cost, $1000 A A 30 A 60 A 90 A 120 A 150 A 180 A 210 A 240
2.31 Apple Computer wants to have $2.1 billion
available 5 years from now to finance production
of a handheld “electronic brain” that, based
on your behavior, will learn how to control
nearly all the electronic devices in your home,
such as the thermostat, coffee pot, TV, and
sprinkler system. The company expects to set
aside uniformly increasing amounts of money
each year to meet its goal. If the amount set
aside at the end of year 1 is $50 million, how
much will the constant increase G have to be
each year? Assume the investment account
grows at a rate of 18% per year.
2.32 Tacozza Electric, which manufactures brush dc
servomotors, budgeted $75,000 per year to pay for
certain components over the next 5 years. If the
company expects to spend $15,000 in year 1, how
much of a uniform (arithmetic) increase each year
is the company expecting in the cost of this part?
Assume the company uses an interest rate of 10%
per year.
Geometric Gradient
2.33 There are no tables in the back of your book for the
geometric gradient series factors. Calculate the first
two annual worth factor values, that is, A values for
n 1 and 2, that would be in a 10% interest table
for a growth rate of 4% per year.
2.34 Determine the present worth of a geometric gradient
series with a cash flow of $50,000 in year 1 and
increases of 6% each year through year 8. The interest
rate is 10% per year.
2.35 Determine the difference in the present worth values
of the following two commodity contracts at
an interest rate of 8% per year.
Contract 1 has a cost of $10,000 in year 1; costs
will escalate at a rate of 4% per year for 10 years.
Contract 2 has the same cost in year 1, but costs
will escalate at 6% per year for 11 years.
2.36 El Paso Water Utilities (EPWU) purchases surface
water for treatment and distribution to EPWU customers
from El Paso County Water Improvement
District during the irrigation season. A new contract
between the two entities resulted in a reduction in
future price increases in the cost of the water from
8% per year to 4% per year for the next 20 years. If
the cost of water next year (which is year 1 of the
new contract) will be $260 per acre-foot, what is
the present worth of the savings (in $/acre-ft) to the
utility between the old and the new contracts? Let
the interest rate equal 6% per year.
2.37 Determine the present worth of a maintenance
contract that has a cost of $30,000 in year 1 and
annual increases of 6% per year for 10 years. Use
an interest rate of 6% per year.
Interest Rate and Rate of Return
2.38 Gesky Industrial Products manufactures brushless
blowers for boilers, food service equipment, kilns,
and fuel cells. The company borrowed $18,000,000
for a plant expansion and repaid the loan in seven
annual payments of $3,576,420, with the first payment
made 1 year after the company received the
money. What was the interest rate on the loan? Use
hand and spreadsheet solutions.
2.39 If the value of Jane’s retirement portfolio increased
from $170,000 to $813,000 over a 1 5-year
period, with no deposits made to the account over
that period, what annual rate of return did she
make?
2.40 A person’s credit score is important in determining
the interest rate on a home mortgage. According to
Consumer Credit Counseling Service, a homeowner
with a $100,000 mortgage and a 520 credit
score will pay $110,325 more in interest charges
over the life of a 30-year loan than a homeowner
with the same mortgage and a credit score of 720.
How much higher would the interest rate per year
have to be in order to account for this much difference
in interest charges, if the $100,000 loan is
repaid in a single lump-sum payment at the end of
30 years?
2.41 During a period when the real estate market in
Phoenix, Arizona, was undergoing a significant
downturn, CSM Consulting Engineers made an
agreement with a distressed seller to purchase an
office building under the following terms: total
price of $1.2 million with a down payment of
$200,000 now and no payments for 4 years, after
which the remaining balance of $1 million would

68 Chapter 2 Factors: How Time and Interest Affect Money
be paid. CSM was able to make this deal because
of poor market conditions at the time of purchase,
and, at the same time, planning to sell the building
in 4 years (when market conditions would probably
be better) and move to a larger office building
in Scottsdale, Arizona. If CSM was able to sell the
building in exactly 4 years for $1.9 million, what
rate of return per year did the company make on
the investment?
2.42 A start-up company that makes hydraulic seals
borrowed $800,000 to expand its packaging and
shipping facility. The contract required the company
to repay the investors through an innovative
mechanism called faux dividends, a series of
uniform annual payments over a fixed period of
time. If the company paid $250,000 per year for
5 years, what was the interest rate on the loan?
2.43 Bessimer Electronics manufactures addressable
actuators in one of its Maquiladora plants in
Mexico. The company believes that by investing
$24,000 each year in years 1, 2, and 3, it will avoid
spending $87,360 in year 3. If the company does
make the annual investments, what rate of return
will it realize?
2.44 UV curable epoxy resins are used in sealing, in gap
filling, and as a clear coating. Your boss saw a report
submitted by the chief financial officer (CFO)
that said the equivalent annual worth of maintaining
the equipment used in producing the resins was
$48,436 over the last 5 years. The report showed
that the cost in year 1 was $42,000, and it increased
arithmetically by $4000 each year. Your boss
thought $48,436 was too high, so she asked you to
determine what interest rate the CFO used in making
the calculations. What was the interest rate?
Number of Years
2.45 Acme Bricks, a masonry products company, wants
to have $600,000 on hand before it invests in new
conveyors, trucks, and other equipment. If the
company sets aside $80,000 per year in an account
that increases in value at a rate of 15% per year,
how many years will it be before Acme can purchase
the equipment?
2.46 An engineer who was contemplating retirement
had $1.6 million in his investment portfolio. However,
a severe recession caused his portfolio to decrease
to only 55% of the original amount, so he
kept working. If he was able to invest his money at
a rate of return of 9% per year after the recession
ended, how many years did it take for his account
to get back to the $1.6 million value?
2.47 You own a small engineering consulting company.
If you invest $200,000 of the company’s money in
a natural gas well that is expected to provide income
of $29,000 per year, how long must the well
produce at that rate in order to get the money back
plus a rate of return of 10% per year?
2.48 A perceptive engineer started saving for her retirement
15 years ago by diligently saving $18,000
each year through the present time. She invested in
a stock fund that averaged a 12% rate of return over
that period. If she makes the same annual investment
and gets the same rate of return in the future,
how long will it be from now (time zero) before she
has $1,500,000 in her retirement fund?
2.49 A mechanical engineering graduate who wanted
to have his own business borrowed $350,000
from his father as start-up money. Because he
was family, his father charged interest at only 4%
per year. If the engineer was able to pay his father
$15,000 in year 1, $36,700 in year 2, and amounts
increasing by $21,700 each year, how many
years did it take for the engineer to repay the
loan?
2.50 The energy costs of a company involved in powder
coating of outdoor furniture are expected to increase
at a rate of $400 per year. The cost at the
end of the next year (year 1) is expected to be
$13,000. How many years will it be from now before
the equivalent annual cost is $16,000 per year,
if interest is 8% per year?
2.51 In cleaning out some files that were left behind by
the engineer who preceded you in your current job,
you found an old report that had a calculation for
the present worth of certain maintenance costs for
state highways. The report contained the following
equation (with cost in $1 million):
12{1 [(1 0.03)(1 0.06)] x } (0.06 0.03)
140
The value of x that was used in the calculation was
illegible. What is its value?
2.52 The equivalent annual worth of an increasing
arithmetic gradient is $135,300. If the cash flow in
year 1 is $35,000 and the gradient amount is
$19,000, what is the value of n at an interest rate of
10% per year?
2.53 You are told that the present worth of an increasing
geometric gradient is $88,146. If the cash flow in
year 1 is $25,000 and the gradient increase is
18% per year, what is the value of n ? The interest
rate is 10% per year.

Additional Problems and FE Exam Review Questions 69
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
2.54 The amount of money that Diamond Systems can
spend now for improving productivity in lieu of
spending $30,000 three years from now at an interest
rate of 12% per year is closest to:
( a) $15,700
( b) $17,800
(c) $19,300
(d ) $21,350
2.55 A manufacturing company spent $30,000 on a new
conveyor belt. If the conveyor belt resulted in cost
savings of $4200 per year, the length of time it
would take for the company to recover its investment
at 8% per year is closest to:
( a) Less than 9 years
( b) 9 to 10 years
( c) 11 to 12 years
( d ) Over 12 years
2.56 Levi Strauss has some of its jeans stone-washed
under a contract with independent U.S. Garment
Corp. If U.S. Garment’s operating cost per machine
is $22,000 for year 1 and increases by a
constant $1000 per year through year 5, what is
the equivalent uniform annual cost per machine
for the 5 years at an interest rate of 8% per year?
( a) $23,850
( b) $24,650
( c) $25,930
( d ) Over $26,000
2.57 The FG factor values can be derived by multiplying:
( a) ( PF ) and ( AG ) factor values
( b) ( FP ) and ( AG ) factor values
( c) ( PF ) and ( PG ) factor values
( d ) ( FP ) and ( PG ) factor values
2.58 At i 4% per year, A for years 1 through 6 of the
cash flows shown below is closest to:
( a) $300
( b) $560
( c) $800
( d ) $1040
0 1
$800
2
$700
3
$600
4
$500
5
$400
6 Years
$300
2.59 The value of the factor ( PF , i ,10) can be found by
getting the factor values for (PF,i,4) and ( PF , i ,6)
and:
(a) Adding the values for ( PF , i ,4) and ( PF , i ,6)
(b) Multiplying the values for ( PF , i ,4) and
( PF , i ,6)
(c) Dividing the value for ( PF , i ,6) by the value
for ( PF , i ,4)
( d ) None of the above
2.60 A small construction company is considering the
purchase of a used bulldozer for $61,000. If the
company purchases the dozer now, the equivalent
future amount in year 4 that the company is
paying for the dozer at 4% per year interest is
closest to:
(a) $52,143
(b) $65,461
(c) $71,365
(d ) Over $72,000
2.61 The cost of lighting and maintaining the tallest
smokestack in the United States (at a shuttered
ASARCO refinery) is $90,000 per year. At an
interest rate of 10% per year, the present worth
of maintaining the smokestack for 10 years is
closest to:
(a) $1,015,000
(b) $894,000
(c) $712,000
(d ) $553,000
2.62 An enthusiastic new engineering graduate plans to
start a consulting firm by borrowing $100,000 at
10% per year interest. The loan payment each year
to pay off the loan in 7 years is closest to:
(a) $18,745
(b) $20,540
(c) $22,960
(d ) $23,450
2.63 An engineer who believed in “save now and play
later” wanted to retire in 20 years with $1.5 million.
At 10% per year interest, to reach the $1.5 million
goal, starting 1 year from now, the engineer must
annually invest:
( a) $26,190
( b) $28,190
( c) $49,350
( d ) $89,680
2.64 The cost of a border fence is $3 million per
mile. If the life of such a fence is assumed to
be 10 years, the equivalent annual cost of a

70 Chapter 2 Factors: How Time and Interest Affect Money
10-mile-long fence at an interest rate of 10%
per year is closest to:
(a) $3.6 million
(b) $4.2 million
(c) $4.9 million
(d) Over $5.0 million
2.65 An investment of $75,000 in equipment that
will reduce the time for machining self-locking
fasteners will save $20,000 per year. At an interest
rate of 10% per year, the number of years
required to recover the initial investment is
closest to:
(a) 6 years
(b) 5 years
(c) 4 years
(d) 3 years
2.66 The number of years required for an account to accumulate
$650,000 if Ralph deposits $50,000 each
year and the account earns interest at a rate of 6%
per year is closest to:
(a) 13 years
(b) 12 years
(c) 11 years
(d) 10 years
2.67 Aero Serve, Inc., manufactures cleaning nozzles
for reverse-pulse jet dust collectors. The company
spent $40,000 on a production control system that
will increase profits by $13,400 per year for
5 years. The rate of return per year on the investment
is closest to:
(a) 20%
(b) 18%
(c) 16%
(d) Less than 15%
2.68 Energy costs for a green chemical treatment have
been increasing uniformly for 5 years. If the cost in
year 1 was $26,000 and it increased by $2000 per
year through year 5, the present worth of the costs
at an interest rate of 10% per year is closest to:
(a) $102,900
(b) $112,300
( c) $122,100
( d) $195,800
2.69 In planning for your retirement, you expect to
save $5000 in year 1, $6000 in year 2, and amounts
increasing by $1000 each year through year 20. If
your investments earn 10% per year, the amount
you will have at the end of year 20 is closest to:
( a) $242,568
(b) $355,407
(c) $597,975
(d) $659,125
2.70 Income from a precious metals mining operation
has been decreasing uniformly for 5 years. If income
in year 1 was $300,000 and it decreased by
$30,000 per year through year 4, the annual worth
of the income at 10% per year is closest to:
(a) $310,500
(b) $258,600
( c) $203,900
( d) $164,800
2.71 If you are able to save $5000 in year 1, $5150 in
year 2, and amounts increasing by 3% each year
through year 20, the amount you will have at the
end of year 20 at 10% per year interest is closest to:
( a) $60,810
( b) $102,250
(c) $351,500
(d) Over $410,000
CASE STUDY
TIME MARCHES ON; SO DOES THE INTEREST RATE
Background
Information
During the last week, Sundara has read about different situations
that involve money, interest rate, and different amounts
of time. She has gotten interested in the major effects that
time and interest rates have on the amount of money necessary
to do things and the significant growth in the amount of
money when a large number of years are considered. In all
cases, the interest focuses on the amount of money at the end
of the time period.
The four situations are described here.
A. Manhattan Island was purchased in 1626 for $24. After
385 years in 2011, at 6% per year compounded interest,
the current value must be very large.
B. At the age of 22, if she saved only $2000 per year for
the next 10 years (starting next year) and made a return
of 6% per year, by today’s standards, she would have
accumulated a nice sum at the age of 70.

Case Study 71
C. A corporation invested $2 million in developing and
marketing a new product in 1945 (just after World
War II, this was a lot of money) and has made a steady
net cash flow of $300,000 per year for some 65 years.
Sundara estimated the annual rate of return must be
quite good, especially given that she is lucky to earn 4%
per year on her own investments these days.
D. A friend who is not good with money, went to a pawn
shop and borrowed $200 for one week and paid $30 in
interest. Sundara thought this might be a pretty good
deal, in case she ever ran low on cash. However, she did
not know whether the interest was simple or compounded
monthly, and how much may be owed were
this loan not paid off for 1 year.
Case Study Exercises
1. What is the annual interest rate for each situation? Include
both the annual simple and the compound rates
for situation D.
2. Calculate and observe the total amount of money involved
in each situation at the end of the time periods
compared to the starting amount. Is the ending amount
larger or smaller than you would expect it to be prior to
making any computations?
3. Think of a situation for yourself that may be similar to
any of those above. Determine the interest rate, the
time period, and the starting and ending amounts of
money.

CHAPTER 3
Combining
Factors and
Spreadsheet
Functions
LEARNING OUTCOMES
Purpose: Use multiple factors and spreadsheet functions to find equivalent amounts for cash flows that have nonstandard
placement.
SECTION TOPIC LEARNING OUTCOME
3.1 Shifted series • Determine the P , F or A values of a series
starting at a time other than period 1.
3.2 Shifted series and single cash
flows
• Determine the P , F , or A values of a shifted series
and randomly placed single cash flows.
3.3 Shifted gradients • Make equivalence calculations for shifted
arithmetic or geometric gradient series that
increase or decrease in size of cash flows.

M
ost estimated cash flow series do not fit exactly the series for which the factors,
equations, and spreadsheet functions in Chapter 2 were developed. For
a given sequence of cash flows, there are usually several correct ways to determine
the equivalent present worth P , future worth F , or annual worth A . This chapter
explains how to combine engineering economy factors and spreadsheet functions to
address more complex situations involving shifted uniform series, gradient series, and
single cash flows.
3.1 Calculations for Uniform Series That Are Shifted
When a uniform series begins at a time other than at the end of period 1, it is called a shifted
series. In this case several methods can be used to find the equivalent present worth P . For
example, P of the uniform series shown in Figure 3–1 could be determined by any of the
following methods:
• Use the P F factor to find the present worth of each disbursement at year 0 and add them.
• Use the F P factor to find the future worth of each disbursement in year 13, add them, and
then find the present worth of the total, using P F ( P F , i ,13).
• Use the F A factor to find the future amount F A ( F A , i ,10), and then compute the present
worth, using P F ( P F , i ,13).
• Use the P A factor to compute the “present worth” P 3 A ( P A , i ,10) (which will be located
in year 3, not year 0), and then find the present worth in year 0 by using the ( P F , i ,3) factor.
Typically the last method is used for calculating the present worth of a uniform series that does
not begin at the end of period 1. For Figure 3–1, the “present worth” obtained using the P A factor
is located in year 3. This is shown as P 3 in Figure 3–2. Note that a P value is always located
1 year or period prior to the beginning of the first series amount. Why? Because the P A factor
was derived with P in time period 0 and A beginning at the end of period 1. The most common
mistake made in working problems of this type is improper placement of P . Therefore, it is extremely
important to remember:
The present worth is always located one period prior to the first uniform series amount when
using the P A factor.
Placement of P
To determine a future worth or F value, recall that the F A factor derived in Section 2.3 had
the F located in the same period as the last uniform series amount. Figure 3–3 shows the location
of the future worth when F A is used for Figure 3–1 cash flows.
The future worth is always located in the same period as the last uniform series amount when
using the F A factor.
It is also important to remember that the number of periods n in the P A or F A factor is equal
to the number of uniform series values. It may be helpful to renumber the cash flow diagram to
avoid errors in counting. Figures 3–2 and 3–3 show Figure 3–1 renumbered to determine n 10.
Placement of F
0 1 2 3 4 5 6 7 8 9 10 11 12 13 Year
Figure 3–1
A uniform series that is
shifted.
A = $50
P 3 =?
0 1 2 3 4 5 6 7 8 9 10 11 12 13 Year
1 2 3 4 5 6 7 8 9 10 n
Figure 3–2
Location of present worth
and renumbering for n for
the shifted uniform series
in Figure 3–1.
A = $50

74 Chapter 3 Combining Factors and Spreadsheet Functions
Figure 3–3
Placement of F and
renumbering for n for the
shifted uniform series of
Figure 3–1.
F =?
0 1 2 3 4 5 6 7 8 9 10 11 12 13 Year
1 2 3 4 5 6 7 8 9 10 n
A = $50
As stated above, several methods can be used to solve problems containing a uniform series
that is shifted. However, it is generally more convenient to use the uniform series factors than the
single-amount factors. Specific steps should be followed to avoid errors:
1. Draw a diagram of the positive and negative cash flows.
2. Locate the present worth or future worth of each series on the cash flow diagram.
3. Determine n for each series by renumbering the cash flow diagram.
4. Draw another cash flow diagram representing the desired equivalent cash flow.
5. Set up and solve the equations.
These steps are illustrated below.
EXAMPLE 3.1
The offshore design group at Bechtel just purchased upgraded CAD software for $5000 now
and annual payments of $500 per year for 6 years starting 3 years from now for annual upgrades.
What is the present worth in year 0 of the payments if the interest rate is 8% per year?
Solution
The cash flow diagram is shown in Figure 3–4. The symbol P A is used throughout this chapter
to represent the present worth of a uniform annual series A , and P ' A represents the present worth
at a time other than period 0. Similarly, P T represents the total present worth at time 0. The
correct placement of P ' A and the diagram renumbering to obtain n are also indicated. Note that
P ' A is located in actual year 2, not year 3. Also, n 6, not 8, for the P A factor. First find the
value of P ' A of the shifted series.
P ' A $500( P A ,8%,6)
Since P ' A is located in year 2, now find P A in year 0.
P A P ' A ( P F ,8%,2)
P T =?
P A =?
1 2
P A =?
i = 8% per year
3 4 5 6 7 8 Year
0 1 2 3 4 5 6
n
A = $500
P 0 = $5000
Figure 3–4
Cash flow diagram with placement of P values, Example 3.1.
The total present worth is determined by adding P A and the initial payment P 0 in year 0.
P T P 0 P A
5000 500( P A ,8%,6)( P F ,8%,2)
5000 500(4.6229)(0.8573)
$6981.60

3.1 Calculations for Uniform Series That Are Shifted 75
The more complex that cash flow series become, the more useful are the spreadsheet functions.
When the uniform series A is shifted, the NPV function is used to determine P , and the
PMT function finds the equivalent A value. The NPV function, like the PV function, determines
the P values, but NPV can handle any combination of cash flows directly from the cells. As we
learned in Chapter 2, enter the net cash flows in contiguous cells (column or row), making sure
to enter “0” for all zero cash flows. Use the format
NPV( i %, second_cell:last_cell) fi rst_cell
First_cell contains the cash flow for year 0 and must be listed separately for NPV to correctly
account for the time value of money. The cash flow in year 0 may be 0.
The easiest way to find an equivalent A over n years for a shifted series is with the PMT function,
where the P value is from the NPV function above. The format is the same as we learned
earlier; the entry for P is a cell reference, not a number.
PMT( i %, n , cell_with_ P , F )
Alternatively, the same technique can be used when an F value was obtained using the FV function.
Now the last entry in PMT is “cell_with_ F .”
It is very fortunate that any parameter in a spreadsheet function can itself be a function. Thus,
it is possible to write the PMT function in a single cell by embedding the NPV function (and FV
function, if needed). The format is
PMT( i %, n , NPV( i %,second_cell:last_cell) first_cell, F ) [3.1]
Of course, the answer for A is the same for the two-cell operation or a single-cell, embedded
function. All three of these functions are illustrated in Example 3.2.
EXAMPLE 3.2
Recalibration of sensitive measuring devices costs $8000 per year. If the machine will be recalibrated
for each of 6 years starting 3 years after purchase, calculate the 8-year equivalent
uniform series at 16% per year. Show hand and spreadsheet solutions.
Solution by Hand
Figure 3–5 a and b shows the original cash flows and the desired equivalent diagram. To convert
the $8000 shifted series to an equivalent uniform series over all periods, first convert the
uniform series into a present worth or future worth amount. Then either the A P factor or the
A F factor can be used. Both methods are illustrated here.
Present worth method. (Refer to Figure 3–5 a .) Calculate P ' A for the shifted series in year 2,
followed by P T in year 0. There are 6 years in the A series.
P ' A 8000( P A ,16%,6)
P T P ' A ( P F ,16%,2) 8000( P A ,16%,6)( P F ,16%,2)
8000(3.6847)(0.7432) $21,907.75
The equivalent series A ' for 8 years can now be determined via the A P factor.
A ' P T ( A P ,16%,8) $5043.60
Future worth method. (Refer to Figure 3–5 a .) First calculate the future worth F in year 8.
F 8000( F A ,16%,6) $71,820
The A F factor is now used to obtain A ' over all 8 years.
A ' F ( A F ,16%,8) $5043.20

76 Chapter 3 Combining Factors and Spreadsheet Functions
P T =? P A =?
F =? i = 16% per year
0 1 2 3 4 5 6 7 8
0 1 2 3 4 5 6 7 8
A = $8000
(a)
A =?
(b)
PMT(16%,8,B12)
PMT(16%,8,NPV(16%,B4:B11) B3
NPV(16%,B4:B11) B3
(c)
Figure 3–5
( a ) Original and ( b ) equivalent cash flow diagrams; and ( c ) spreadsheet functions to determine
P and A , Example 3.2.
Solution by Spreadsheet
(Refer to Figure 3–5 c .) Enter the cash flows in B3 through B11 with entries of “0” in the first
three cells. Use the NPV function to display P $21,906.87.
There are two ways to obtain the equivalent A over 8 years. Of course, only one of these
PMT functions needs to be entered. (1) Enter the PMT function making direct reference to
the P value (see cell tag for DE5), or (2) use Equation [3.1] to embed the NPV function into
the PMT function (see cell tag for DE8).
3.2 Calculations Involving Uniform Series and Randomly
Placed Single Amounts
When a cash flow includes both a uniform series and randomly placed single amounts, the procedures
of Section 3.1 are applied to the uniform series and the single-amount formulas are applied
to the one-time cash flows. This approach, illustrated in Examples 3.3 and 3.4, is merely a
combination of previous ones. For spreadsheet solutions, it is necessary to enter the net cash
flows before using the NPV and other functions.
EXAMPLE 3.3
An engineering company in Wyoming that owns 50 hectares of valuable land has decided to
lease the mineral rights to a mining company. The primary objective is to obtain long-term
income to finance ongoing projects 6 and 16 years from the present time. The engineering
company makes a proposal to the mining company that it pay $20,000 per year for 20 years
beginning 1 year from now, plus $10,000 six years from now and $15,000 sixteen years from
now. If the mining company wants to pay off its lease immediately, how much should it pay
now if the investment is to make 16% per year?

3.2 Calculations Involving Uniform Series and Randomly Placed Single Amounts 77
Solution
The cash flow diagram is shown in Figure 3–6 from the owner’s perspective. Find the present
worth of the 20-year uniform series and add it to the present worth of the two one-time amounts
to determine P T' .
P T 20,000( P A ,16%,20) 10,000( P F ,16%,6) 15,000( P F ,16%,16)
$124,075
Note that the $20,000 uniform series starts at the end of year 1, so the P A factor determines the
present worth at year 0.
$10,000
$15,000
A = $20,000
0 1 2 3 4 5 6 7 15 16 17 18 19 20 Year
P T = ?
i = 16% per year
Figure 3–6
Diagram including a uniform series and single amounts, Example 3.3.
When you calculate the A value for a cash flow series that includes randomly placed single
amounts and uniform series, first convert everything to a present worth or a future worth.
Then you obtain the A value by multiplying P or F by the appropriate A P or A F factor.
Example 3.4 illustrates this procedure.
EXAMPLE 3.4
A design-build-operate engineering company in Texas that owns a sizable amount of land
plans to lease the drilling rights (oil and gas only) to a mining and exploration company. The
contract calls for the mining company to pay $20,000 per year for 20 years beginning 3 years
from now (i.e., beginning at the end of year 3 and continuing through year 22) plus $10,000 six
years from now and $15,000 sixteen years from now. Utilize engineering economy relations by
hand and by spreadsheet to determine the five equivalent values listed below at 16% per year.
1. Total present worth P T in year 0
2. Future worth F in year 22
3. Annual series over all 22 years
4. Annual series over the first 10 years
5. Annual series over the last 12 years
Solution by Hand
Figure 3–7 presents the cash flows with equivalent P and F values indicated in the correct years
for the P A , P F , and F A factors.
1. P T in year 0: First determine P' A of the series in year 2. Then P T is the sum of three
P values: the series present worth value moved back to t 0 with the P F factor, and the
two P values at t 0 for the two single amounts in years 6 and 16.

78 Chapter 3 Combining Factors and Spreadsheet Functions
P ' A 20,000( P A ,16%,20)
P T P ' A ( P F ,16%,2) 10,000( P F ,16%,6) 15,000( P F ,16%,16)
20,000( P A ,16%,20)( P F ,16%,2) 10,000( P F ,16%,6)
15,000( P F ,16%,16)
$93,625 [3.2]
2. F in year 22: To determine F in year 22 from the original cash flows (Figure 3–7), find
F for the 20-year series and add the two F values for the two single amounts. Be sure to
carefully determine the n values for the single amounts: n 22 6 16 for the $10,000
amount and n 22 16 6 for the $15,000 amount.
F 20,000( F A ,16%,20) 10,000( F P ,16%,16) 15,000( F P ,16%,6) [3.3]
$2,451,626
3. A over 22 years: Multiply P T $93,625 from (1) above by the A P factor to determine
an equivalent 22-year A series, referred to as A 1–22 here.
A 1–22 P T ( A P ,16%,22) 93,625(0.16635) $15,575 [3.4]
An alternate way to determine the 22-year series uses the F value from (2) above. In this
case, the computation is A 1–22 F ( A F ,16%,22) $15,575.
$15,000
$10,000
$20,000
0 1 2 3 4 5 13 20 n
0 1 2 3 4 5 6 7 15 16 17 18 19 20 21 22 Year
P T =?
P A =?
Figure 3–7
Diagram for Example 3.4.
i = 16% per year
F =?
4. A over years 1 to 10: This and (5), which follows, are special cases that often occur in
engineering economy studies. The equivalent A series is calculated for a number of years
different from that covered by the original cash flows. This occurs when a defined study
period or planning horizon is preset for the analysis. (More is mentioned about study periods
later.) To determine the equivalent A series for years 1 through 10 only (call it A 1–10 ),
the P T value must be used with the A P factor for n 10. This computation transforms
Figure 3–7 into the equivalent series A 1–10 in Figure 3–8 a .
A 1–10 P T ( A P ,16%,10) 93,625(0.20690) $19,371 [3.5]
5. A over years 11 to 22: For the equivalent 12-year series for years 11 through 22 (call it
A 11–22 ), the F value must be used with the A F factor for 12 years. This transforms Figure
3–7 into the 12-year series A 11–22 in Figure 3–8 b .
A 11–22 F ( A F ,16%,12) 2,451,626(0.03241) $79,457 [3.6]
Notice the huge difference of more than $60,000 in equivalent annual amounts that occurs
when the present worth of $93,625 is allowed to compound at 16% per year for the first
10 years. This is another demonstration of the time value of money.
Solution by Spreadsheet
Figure 3–9 is a spreadsheet image with answers for all five questions. The $20,000 series and
the two single amounts have been entered into separate columns, B and C. The zero cash flow

P T = $93,625 A 1–10 = ? (a)
3.2 Calculations Involving Uniform Series and Randomly Placed Single Amounts 79
1 2 3 4 5 6 7 8 9 10 11 21 22
Year
Figure 3–8
Cash flows of Figure 3–7
converted to equivalent
uniform series for
( a ) years 1 to 10 and
( b ) years 11 to 22.
i = 16%
A 11–22 = ? Figure 3–9
0 1 9 10 11 12 13 14
15 16 17 18 19 20 21 22 Year
i = 16%
(b)
F = $2,451,626
Spreadsheet using cell
reference format.
Example 3.4.
values are all entered so that the functions will work correctly. This is an excellent example
demonstrating the versatility of the NPV, FV, and PMT functions. To prepare for sensitivity
analysis, the functions are developed using cell reference format or global variables, as indicated
in the column E function. This means that virtually any number—the interest rate, any
cash flow estimate in the series or the single amounts, and the timing within the 22-year time
frame—can be changed and the new answers will be immediately displayed.
1. Present worth values for the series and single amounts are determined in cells F6 and F10,
respectively, using the NPV function. The sum of these in F14 is P T $93,622, which
corresponds to the value in Equation [3.2]. Alternatively, P T can be determined directly
via the sum of two NPV functions, shown in row 15.
2. The FV function in row 18 uses the P value in F14 (preceded by a minus sign) to determine
F twenty-two years later. This is significantly easier than Equation [3.3].
3. To find the 22-year A series starting in year 1, the PMT function in row 21 references
the P value in cell F14. This is effectively the same procedure used in Equation [3.4] to
obtain A 1–22 .

80 Chapter 3 Combining Factors and Spreadsheet Functions
For the spreadsheet enthusiast, it is possible to find the 22-year A series value directly
by using the PMT function with embedded NPV functions. The cell reference format is
PMT(D1,22,−(NPV(D1,B6:B27)B5 NPV(D1,C6:C27)C5)).
4. and 5. It is quite simple to determine an equivalent uniform series over any number of
periods using a spreadsheet, provided the series starts one period after the P value is located
or ends in the same period that the F value is located. These are both true for the
series requested here. The results in F24 and F27 are the same as A 1–10 and A 11–22 in Equations
[3.5] and [3.6], respectively.
Comment
Remember that round-off error will always be present when comparing hand and spreadsheet
results. The spreadsheet functions carry more decimal places than the tables during calculations.
Also, be very careful when constructing spreadsheet functions. It is easy to miss a value,
such as the P or F in PMT and FV functions, or a minus sign between entries. Always check
your function entries carefully before touching .
3.3 Calculations for Shifted Gradients
In Section 2.5, we derived the relation P G ( P G , i , n ) to determine the present worth of the
arithmetic gradient series. The P G factor, Equation [2.25], was derived for a present worth in
year 0 with the gradient first appearing in year 2.
Placement of gradient P
The present worth of an arithmetic gradient will always be located two periods before the
gradient starts.
Refer to Figure 2–14 as a refresher for the cash flow diagrams.
The relation A G ( A G , i , n ) was also derived in Section 2.5. The A G factor in Equation
[2.27] performs the equivalence transformation of a gradient only into an A series from
years 1 through n (Figure 2–15). Recall that the base amount must be treated separately. Then the
equivalent P or A values can be summed to obtain the equivalent total present worth P T and total
annual series A T .
A conventional gradient series starts between periods 1 and 2 of the cash flow sequence. A
gradient starting at any other time is called a shifted gradient . The n value in the P G and A G
factors for a shifted gradient is determined by renumbering the time scale. The period in which
the gradient fi rst appears is labeled period 2. The n value for the gradient factor is determined by
the renumbered period where the last gradient increase occurs.
Partitioning a cash flow series into the arithmetic gradient series and the remainder of the cash
flows can make very clear what the gradient n value should be. Example 3.5 illustrates this partitioning.
EXAMPLE 3.5
Fujitsu, Inc. has tracked the average inspection cost on a robotics manufacturing line for
8 years. Cost averages were steady at $100 per completed unit for the first 4 years, but have
increased consistently by $50 per unit for each of the last 4 years. Analyze the gradient increase,
using the P G factor. Where is the present worth located for the gradient? What is the
general relation used to calculate total present worth in year 0?
Solution
The cash flow diagram in Figure 3–10 a shows the base amount A $100 and the arithmetic
gradient G $50 starting between periods 4 and 5. Figure 3–10 b and c partitions these two
series. Gradient year 2 is placed in actual year 5 of the entire sequence in Figure 3–10 c . It is
clear that n 5 for the P G factor. The P G ? arrow is correctly placed in gradient year 0,
which is year 3 in the cash flow series.

3.3 Calculations for Shifted Gradients 81
0 1 2 3 4 5 6 7 8 Year
A = $100
G = $50
(a)
$150
$200
$250
$300
0 1 2 3 4 5 6 7 8 Year
P G = ?
A = $100
(b)
0 1 2
3 4 5 6 7 8 Year
Figure 3–10
Partitioned cash flow, ( a ) ( b ) ( c ), Example 3.5.
0 1 2 3 4 5
G = $50
(c)
The general relation for P T is taken from Equation [2.19]. The uniform series A $100
occurs for all 8 years, and the G $50 gradient present worth P G appears in year 3.
$50
$100
$150
P T P A P G 100( P A , i ,8) 50( P G , i ,5)( P F , i ,3)
$200
Gradient
n
It is important to note that the A G factor cannot be used to find an equivalent A value in periods
1 through n for cash flows involving a shifted gradient. Consider the cash flow diagram of
Figure 3–11. To find the equivalent annual series in years 1 through 10 for the gradient series
only, first find the present worth P G of the gradient in actual year 5, take this present worth back
to year 0, and annualize the present worth for 10 years with the A P factor. If you apply the annual
series gradient factor ( A G , i ,5) directly, the gradient is converted into an equivalent annual
series over years 6 through 10 only, not years 1 through 10, as requested.
To find the equivalent A series of a shifted gradient through all the n periods, first find the
present worth of the gradient at actual time 0, then apply the (AP, i, n) factor.
P G =?
0 1 2 3 4 5 6 7 8 9 10 Year
$10
$10
$10
$10
0
$10
Figure 3–11
Determination of G and n values used in factors for a shifted gradient.
1
$50
2
$65
3
$80
4
$95
5 Gradient n
$110
G = $15

82 Chapter 3 Combining Factors and Spreadsheet Functions
EXAMPLE 3.6
Set up the engineering economy relations to compute the equivalent annual series in years 1
through 7 for the cash flow estimates in Figure 3–12.
0 1 2 3 4 5 6 7
$50 $50 $50
$70
Figure 3–12
Diagram of a shifted gradient, Example 3.6.
$90
$110
$130
Solution
The base amount annual series is A B $50 for all 7 years (Figure 3–13). Find the present worth
P G in actual year 2 of the $20 gradient that starts in actual year 4. The gradient n is 5.
P G 20( P G , i ,5)
Bring the gradient present worth back to actual year 0.
P 0 P G ( P F , i ,2) 20( P G , i ,5)( P F , i ,2)
Annualize the gradient present worth from year 1 through year 7 to obtain A G .
A G P 0 ( A P , i ,7)
Finally, add the base amount to the gradient annual series.
A T 20( P G , i ,5)( P F , i ,2)( A P , i ,7) 50
For a spreadsheet solution, enter the original cash flows into adjacent cells, say, B3 through
B10, and use an embedded NPV function in the PMT function. The single-cell function is
PMT( i %,7,−NPV( i %, B3:B10)).
P 0 =? P G =?
A T =?
0 1 2 3 4 5 6 7
0 1 2 3 4 5
Year
Gradient n
$50 $50 $50
A B = $50
$70
$90
G = $20
$110
Figure 3–13
Diagram used to determine A for a shifted gradient, Example 3.6.
$130
In Section 2.6, we derived the relation P g A 1 ( PA,g,i,n ) to determine the present worth of a
geometric gradient series, including the initial amount A 1 . The factor was derived to find the
present worth in year 0, with A 1 in year 1 and the first gradient appearing in year 2.
Placement of
gradient P
The present worth of a geometric gradient series will always be located two periods before
the gradient stars, and the initial amount is included in the resulting present worth. Refer
to Figure 2–21 as a refresher for the cash flows.
Equation [2.35] is the formula used for the factor. It is not tabulated.

3.3 Calculations for Shifted Gradients 83
EXAMPLE 3.7
Chemical engineers at a Coleman Industries plant in the Midwest have determined that a
small amount of a newly available chemical additive will increase the water repellency of
Coleman’s tent fabric by 20%. The plant superintendent has arranged to purchase the additive
through a 5-year contract at $7000 per year, starting 1 year from now. He expects
the annual price to increase by 12% per year thereafter for the next 8 years. Additionally,
an initial investment of $35,000 was made now to prepare a site suitable for the contractor
to deliver the additive. Use i 15% per year to determine the equivalent total present
worth for all these cash flows.
Solution
Figure 3–14 presents the cash flows. The total present worth P T is found using g 0.12 and
i 0.15. Equations [2.34] and [2.35] are used to determine the present worth P g for the entire
geometric series at actual year 4, which is moved to year 0 using ( P F ,15%,4).
P T 35,000 A ( P A ,15%,4) A 1 ( P A ,12%,15%,9)( P F ,15%,4)
35,000 7000(2.8550) [ 7000 1 (1.121.15) 9
———————
0.15 0.12 ] (0.5718)
35,000 19,985 28,247
$83,232
Note that n 4 in the ( P A ,15%,4) factor because the $7000 in year 5 is the cash flow of the
initial amount A 1 .
For solution by spreadsheet, enter the cash flows of Figure 3–14. If cells B1 through B14
are used, the function to find P $83,230 is
NPV(15%,B2:B14)B1
The fastest way to enter the geometric series is to enter $7000 for year 5 (into cell B6) and set
up each succeeding cell multiplied by 1.12 for the 12% increase.
P T =? P g =? i = 15% per year
0 1 2 3 4 5 6 7 8 9 10 11 12 13 Year
$7000
0 1 2 3 4 5 6 7 8 9
$7840
Geometric
gradient n
$35,000
12% increase
per year
$17,331
Figure 3–14
Cash flow diagram including a geometric gradient with g 12%, Example 3.7.
Decreasing arithmetic and geometric gradients are common, and they are often shifted gradient
series . That is, the constant gradient is – G or the percentage change is – g from one period to
the next, and the first appearance of the gradient is at some time period (year) other than year 2
of the series. Equivalence computations for present worth P and annual worth A are basically the
same as discussed in Chapter 2, except for the following.

84 Chapter 3 Combining Factors and Spreadsheet Functions
$800
$700
$800
$600
$500
$400
0 1 2 3 4 5
0 1 2 3 4 5 6
Gradient n
Year
P T =?
(a)
$800
A B = $800
G = $100
$400
$300
$200
$100
0 1 2 3 4 5 6
Year
–
0 1 2 3 4 5 6 Year
P A =?
(b)
P G =?
(c)
Figure 3–15
Partitioned cash flow of a shifted arithmetic gradient, (a) (b) (c).
For shifted, decreasing gradients:
• The base amount A (arithmetic) or initial amount A 1 (geometric) is the largest amount in the
first year of the series.
• The gradient amount is subtracted from the previous year’s amount, not added to it.
• The amount used in the factors is –G for arithmetic and –g for geometric gradient series.
• The present worth P G or P g is located 2 years prior to the appearance of the first gradient;
however, a PF factor is necessary to find the present worth in year 0.
Figure 3–15 partitions a decreasing arithmetic gradient series with G $−100 that is shifted
1 year forward in time. P G occurs in actual year 1, and P T is the sum of three components.
P T $800( P F , i ,1) 800( P A , i ,5)( P F , i ,1) 100( P G , i ,5)( P F , i ,1)
EXAMPLE 3.8
Morris Glass Company has decided to invest funds for the next 5 years so that development of
“smart” glass is well funded in the future. This type of new-technology glass uses electrochrome
coating to allow rapid adjustment to sun and dark in building glass, as well as assisting
with internal heating and cooling cost reduction. The financial plan is to invest first, allow appreciation
to occur, and then use the available funds in the future. All cash flow estimates are
in $1000 units, and the interest rate expectation is 8% per year.
Years 1 through 5: Invest $7000 in year 1, decreasing by $1000 per year through year 5.
Years 6 through 10: No new investment and no withdrawals.
Years 11 through 15: Withdraw $20,000 in year 11, decreasing 20% per year through year 15.

3.3 Calculations for Shifted Gradients 85
Determine if the anticipated withdrawals will be covered by the investment and appreciation
plans. If the withdrawal series is over- or underfunded, what is the exact amount available in
year 11, provided all other estimates remain the same?
Solution by Hand
Figure 3–16 presents the cash flow diagram and the placement of the equivalent P values used
in the solution. Calculate the present worth of both series in actual year 0 and add them to determine
if the investment series is adequate to fund the anticipated withdrawals.
P g,0 =?
P g,10 =?
$20,000
A 1 = $20,000
g = 0.20
i = 8% per year
$8192
0 1 2 3 4 5 0 1 2 3 4 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Gradient n
Year
P G =?
A = $7000
G = $1000
Figure 3–16
Investment and withdrawal series, Example 3.8.
Investment series: Decreasing, conventional arithmetic series starting in year 2 with A
$7000, G $1000, and gradient n 5 years. The present worth in year 0 is
P G [7000(PA,8%,5) 1000(PG,8%,5)]
$20,577
Withdrawal series: Decreasing, shifted geometric series starting in year 12 with A 1 $20,000,
g 0.20, and gradient n 5 years. If the present worth in year 10 is identified as P g,10 , the
present worth in year 0 is P g,0 . Use Equation [2.35] for the (PA,20%,8%,5) factor.
The net total present worth is
P g,0 P g,10 (PF,i,n) A 1 (PA,g,i,n)(PF,i,n) [3.7]
20,000 { 1 [——————
1 (0.20)
1 0.08 ] 5
————————
0.08 (0.20)
} (0.4632)
20,000(2.7750)(0.4632)
$25,707
P T 20,577 25,707 $5130
This means that more is withdrawn than the investment series earns. Either additional funds
must be invested or less must be withdrawn to make the series equivalent at 8% per year.
To find the exact amount of the initial withdrawal series to result in P T 0, let A 1 be an
unknown in Equation [3.7] and set P g,0 P G 20,577.
20,577 A 1 (2.7750)(0.4632)
A 1 $16,009 in year 11
The geometric series withdrawal would be 20% less each year.

86 Chapter 3 Combining Factors and Spreadsheet Functions
NPV(8%,B4:B18) + B3
(a)
(b)
Figure 3–17
Spreadsheet solution, Example 3.8. (a) Cash flows and NPV function and (b) Goal Seek to determine initial withdrawal
amount in year 11.
Solution by Spreadsheet
See Figure 3–17a. To determine if the investment series will cover the withdrawal series, enter
the cash flows (in column B using the functions shown in column C) and apply the NPV function
shown in the cell tag to display P T $5130 directly. As above, the sign indicates that,
from a time value of money perspective, there is more withdrawn than invested and earned.
The Goal Seek tool is very handy in determining the initial withdrawal amount that results
in P T 0 (cell B19). Figure 3–17b shows the template and result A 1 $16,009 in year 11.
Each succeeding withdrawal is 80% of the previous one.
Comment
If the withdrawal series is fixed as estimated initially and the investment series base amount A
can be increased, the Goal Seek tool can be used to, again, set P T 0 (cell B19). However, now
establish the entry in B4 as the changing cell. The response is A $8285, and, as before,
succeeding investments are $1000 less.
CHAPTER SUMMARY
In Chapter 2, we derived the equations to calculate the present, future, or annual worth of specific
cash flow series. In this chapter, we learned how to use these equations on cash flow series that
are shifted in time from those for which the basic relations are derived. For example, when a
uniform series does not begin in period 1, we still use the P A factor to find the “present worth”
of the series, except the P value is located one period ahead of the first A value, not at time 0. For
arithmetic and geometric gradients, the P value is two periods ahead of where the gradient starts.
With this information, it is possible to solve for P , A , or F for any conceivable cash flow series.
We have used the power of spreadsheet functions in determining P , F , and A values by singlecell
entries and using cash flow estimates entered into a series of spreadsheet cells. Though
spreadsheet solutions are fast, they do remove some of the understanding of how the time value
of money and the factors change the equivalent value of money.
PROBLEMS
Present Worth Calculations
3.1 Industrial Electric Services has a contract with the
U.S. Embassy in Mexico to provide maintenance
for scanners and other devices in the building. If
the first payment of $12,000 is received now ,
what is the present worth of the contract, provided
the company will receive a total of 10 payments
(i.e., years 0 through 9) and the interest rate is
10% per year?

Problems 87
3.2 Civil engineering consulting firms that provide services
to outlying communities are vulnerable to a
number of factors that affect the financial condition
of the communities, such as bond issues and real
estate developments. A small consulting firm entered
into a fixed-price contract with a large developer,
resulting in a stable income of $260,000 per
year in years 1 through 3. At the end of that time, a
mild recession slowed the development, so the parties
signed another contract for $190,000 per year
for 2 more years. Determine the present worth of the
two contracts at an interest rate of 10% per year.
3.3 The net cash flow associated with development
and sale of a new product is shown. Determine the
present worth at an interest rate of 12% per year.
The cash flow is in $1000 units. Show ( a ) hand and
( b ) spreadsheet solutions.
Year 1 2 3 4 5 6 7 8 9
Cash
Flow, $ −120 −100 −40 50 50 80 80 80 80
0
3.4 Standby power for water utility pumps and other
electrical devices is provided by diesel-powered
generators. As an alternative, the utility can use
natural gas to power the generators, but it will be a
few years before the gas is available at remote
sites. The utility estimates that by switching to gas,
it will save $22,000 per year, starting 3 years from
now. At an interest rate of 8% per year, determine
the present worth in year 0 of the projected savings
that will occur in years 3 through 10.
3.5 Find the present worth at i 10% per year for the
cash flow series shown below.
1
i = 10% per year
2 3 4 5 6 7 8
$200 $200
$200
$90 $90 $90
Year
3.6 The subsidized cost of producing water at the El
Paso Water Utilities (EPWU) Kay Bailey Hutchison
(KBH) desalting plant is $1.56 per 1000 gallons.
Under a contract that EPWU has with Fort Bliss,
the utility sells water to the Army post at a discounted
price of $1.28 per 1000 gallons. (The
KBH plant was built on Army property.) If Fort
Bliss uses 2 billion gallons of water each year,
what is the present worth of the discount for a 20-
year period at an interest rate of 6% per year?
3.7 The rising cost of athletic programs at major universities
has induced college presidents and athletic
directors to devise innovative ways to finance cashstrapped
sports programs. One of the latest schemes
for big-time athletics is the “sports mortgage.” At
the University of Kansas, Jayhawk fans can sign up
to pay $105,000 now, or over a 10-year period, for
the right to buy top seats for football games during
the next 30 years. In return, the seats themselves
will stay locked in at current-year prices. Season
tickets in tier 1 are currently selling for $350 each.
A fan plans to purchase the sports mortgage along
with a current-season ticket and pay for both now ,
then buy a ticket each year for the next 30 years.
What is the total present worth of the pricing plan at
an interest rate of 10% per year?
3.8 The cash flow associated with making self-locking
fasteners is shown below. Determine the net present
worth (year 0) at an interest rate of 10% per
year.
Year 0 1 2 3 4 5 6 7 8 9
Income, $1000 20 20 20 20 30 30 30 30 30 30
Cost, $1000 8 8 8 8 12 12 12 12 12 25
3.9 Beckman Technologies, a relatively small manufacturer
of precision laboratory equipment, borrowed
$2 million to renovate one of its testing
labs. In an effort to pay off the loan quickly, the
company made four payments in years 1 through
4, with each payment being twice as large as the
preceding one. At an interest rate of 10% per year,
what was the size of the first payment?
Annual Worth Calculations
3.10 Revenue from the sale of ergonomic hand tools
was $300,000 in years 1 through 4 and $465,000 in
years 5 through 9. Determine the equivalent annual
revenue in years 1 through 9 at an interest rate
of 10% per year.
3.11 Two engineering graduates who recently got married
are planning for their early retirement 20 years
from now. They believe that they will need
$2,000,000 in year 20. Their plan is to live on one
of their salaries and invest the other. They already
have $25,000 in their investment account. ( a ) How
much will they have to invest each year if the account
grows at a rate of 10% per year? ( b ) If the
maximum they have available to invest each year
is $40,000, will they reach their goal of $2 million
by year 20?
3.12 Costs associated with the manufacture of miniature
high-sensitivity piezoresistive pressure transducers
are $73,000 per year. A clever industrial
engineer found that by spending $16,000 now to
reconfigure the production line and reprogram

88 Chapter 3 Combining Factors and Spreadsheet Functions
two of the robotic arms, the cost will go down to
$58,000 next year and $52,000 in years 2 through 5.
Using an interest rate of 10% per year, determine
( a ) the equivalent annual cost of the manufacturing
operations and (b) the equivalent annual savings in
years 1 through 5.
3.13 Calculate the equivalent annual cost in years 1
through 9 of the following series of disbursements.
Use an interest rate of 10% per year. Show ( a )
hand and ( b ) spreadsheet solutions.
Year Disbursement, $ Year Disbursement, $
0 8000 5 4000
1 4000 6 5000
2 4000 7 5000
3 4000 8 5000
4 4000 9 5000
3.14 For the cash flows below, find the value of x that
makes the equivalent annual worth in years 1
through 7 equal to $300 per year. Use an interest
rate of 10% per year. Show solutions ( a ) by hand
and ( b ) using the Goal Seek tool.
Year Cash Flow, $ Year Cash Flow, $
0 200 4 x
1 200 5 200
2 200 6 200
3 200 7 200
3.15 Precision Instruments, Inc. manufactures highsensitivity
mini accelerometers designed for
modal analysis testing. The company borrowed
$10,000,000 with the understanding that it would
make a $2,000,000 payment at the end of year 1
and then make equal annual payments in years 2
through 5 to pay off the loan. If the interest rate on
the loan was 9% per year, how much was each
payment in years 2 through 5?
3.16 A construction management company is examining
its cash flow requirements for the next 7 years. The
company expects to replace office machines and
computer equipment at various times over the
7-year planning period. Specifically, the company
expects to spend $6000 one year from now, $9000
three years from now, and $10,000 six years from
now. What is the annual worth (in years 1 through 7)
of the planned expenditures, at an interest rate of
10% per year?
3.17 Find the equivalent annual worth for the cash
flows shown, using an interest rate of 12% per
year. Monetary units are $1000.
Year 1 2 3 4 5 6 7 8 9
Cash Flow, $ 20 20 20 20 60 60 60 60 60
3.18 Calculate the net annual worth in years 1 through
10 of the following series of incomes and expenses,
if the interest rate is 10% per year.
Year Income, $Year Expense, $Year
0 0 2500
1–4 700 200
5–10 2000 300
3.19 The city of El Paso gave the El Paso Tennis and
Swim Club a 20-year lease to continue using a
10-acre arroyo park for its facilities. The club
will pay $1000 per year plus make $350,000 in
improvements at the park. In addition, the club
will open its tennis courts to the public from 1
to 5 P.M. Monday through Thursday. If the club
makes $100,000 worth of improvements now
and then $50,000 worth each year for the next
5 years, what is the equivalent annual cost of
the lease to the club at an interest rate of 10%
per year?
3.20 Stadium Capital Financing Group is a Chicago
company that conceived the so-called sports
mortgage wherein fans agree to pay a relatively
large amount of money over a 10- to 30-year
period for the right to buy top seats for football
games for up to the next 50 years. In return,
season ticket prices stay locked in at currentyear
prices, and the package can be sold in the
secondary market, while taking a tax write-off
for donating to a school. Assume a fan buys a
sports mortgage at West Virginia University for
$150,000 that is to be paid over a 10-year period
with the right to buy two season tickets for
$300 each for the next 30 years. The first
payment is made now (i.e., beginning-of-year
payment), and an additional nine payments are
to be made at the end of each year for the next
9 years. Assume the fan also purchases the two
season tickets (also beginning-of-year payments).
What is the total amount of the payment
each year in years 0 through 9 ? Use an interest
rate of 10% per year.
Future Worth Calculations
3.21 The expansion plans of Acme Granite, Stone &
Brick call for the company to add capacity for a
new product in 5 years. The company wants to
have $360,000 available before it announces the
product. If the company sets aside $55,000 now
and $90,000 in year 2, what uniform annual
amount will it have to put in an account in years 3
through 5 to have the $360,000? Assume the account
earns interest at 8% per year.

Problems 89
3.22 For the cash flows shown, calculate the future
worth in year 8 using i 10% per year.
0
1
i = 10% per year
2 3 4 5 6 7 8
Year
Year 0 1 2 3 4 5 6
Cash Flow, $ 100 100 100 100 100 300 300
3.23 Assume you plan to start an annuity plan by making
your first deposit now . If you make annual
deposits of a uniform amount A into the account
that earns interest at a rate of 7% per year, how
many years from now will it be until the value of
the account is equal to 10 times the value of a
single deposit?
3.24 New actuator element technology enables engineers
to simulate complex computer-controlled
movements in any direction. If the technology results
in cost savings in the design of amusement
park rides, what is the future worth in year 5 of savings
of $70,000 now and $20,000 per year in years
1 through 3 at an interest rate of 10% per year?
3.25 Austin Utilities is planning to install solar panels
to provide some of the electricity for its groundwater
desalting plant. The project would be done in
two phases. The first phase will cost $4 million in
year 1 and $5 million in year 2. This investment
will result in energy savings (phase 2) of $540,000
in year 3, $546,000 in year 4, and amounts increasing
by $6000 each year through year 10. Let i
10% per year.
( a) What is the future worth of the savings ?
( b) Is the cost of the solar project justified by the
savings? (Hint: Calculate the difference between
savings and cost).
3.26 For the cash flow diagram shown, determine the
value of W that will render the equivalent future
worth in year 8 equal to $500 at an interest rate of
10% per year.
0
1
i = 10% per year
2 3 4 5 6 7 8
Year
x
x
x
x
Random Single Amounts and Uniform Series
3.28 A small oil company is planning to replace its
Coriolis flowmeters with Emerson Hastelloy flowmeters.
The replacement process will cost the
company $50,000 three years from now. How
much money must the company set aside each year
beginning now (year 0) in order to have the total
amount available immediately after making the
last deposit at the end of year 3? Assume the company
can invest its funds at 15% per year.
3.29 A company that manufactures air-operated drain
valve assemblies budgeted $74,000 per year to pay
for plastic components over a 5-year period. If the
company spent only $42,000 in year 1, what uniform
annual amount should the company expect to
spend in each of the next 4 years to expend the
entire budget? Assume the company uses an interest
rate of 10% per year.
3.30 A recently hired chief executive officer wants to
reduce future production costs to improve the company’s
earnings, thereby increasing the value of the
company’s stock. The plan is to invest $40,000
now and $40,000 in each of the next 2 years to improve
productivity. By how much must annual
costs decrease in years 3 through 7 to recover the
investment plus a return of 12% per year?
3.31 Use the cash flow diagram to determine the single
amount of money Q 4 in year 4 that is equivalent to
all of the cash flows shown. Use i 10% per year.
Q 4 =?
x
2x
2x
2x
i = 10% per year
$40 $40 $40 $40 $40
$40 $40
1
0
1
2 3 4 5 6 7 8
Year
W
3.27 For the cash flows shown in the diagram, determine
the value of x that will make the future worth
in year 8 equal to $70,000.
$25 $25 $25 $25 $25
$25
$25
$50
$50 $50

90 Chapter 3 Combining Factors and Spreadsheet Functions
3.32 For the following series of income and expenses,
find the equivalent value in year 9 at an interest rate
of 12% per year. Show ( a ) hand and ( b ) spreadsheet
solutions.
Years Income, $ Expense, $
0 0 70,000
1–6 9,000 13,000
7–9 28,000 14,000
10–16 38,000 19,000
3.33 An investor just purchased property under a unique
financing agreement with the seller. The contract
price is $1.6 million. The payment plan is Z dollars
now, 2 Z dollars in year 2, and 3 Z dollars in years 3
through 5. If the interest rate on the transaction is
10% per year, how much is the payment in year 2?
3.34 A foursome of entrepreneurial electrical engineering
graduates has a plan to start a new solar power
equipment company based on STE (solar thermal
electric) technology. They have recently approached
a group of investors with their idea and asked for a
loan of $5 million. Within the agreement, the loan is
to be repaid by allocating 80% of the company’s
profits each year for the first 4 years to the investors.
In the fifth year, the company will pay the balance
remaining on the loan in cash. The company’s business
plan indicates that they expect to make no
profit for the first year, but in years 2 through 4, they
anticipate profits to be $1.5 million per year. If the
investors accept the deal at an interest rate of 15%
per year, and the business plan works to perfection,
what is the expected amount of the last loan payment
(in year 5 )?
Shifted Gradients
3.35 Find the present worth in year 0 for the cash flows
shown. Let i 10% per year.
0
1
$50
$50
i = 10% per year
2 3 4 5 6 7 8
$70
$90
$170
$130
$150
$170
Year
3.36 For the cash flows shown, determine the present
worth in year 0, if the interest rate is 12% per year.
Year 1 2 3 4 5 6 7 8 9 10
Cash Flow, $ 13 13 13 13 16 19 22 25 28 31
3.37 A low-cost noncontact temperature measuring tool
may be able to identify railroad car wheels that are in
need of repair long before a costly structural failure
occurs. If BNF Railroad saves $100,000 in years 1
through 5, $110,000 in year 6, and constant amounts
increasing by $10,000 each year through year 20,
what is the equivalent annual worth over the 20 years
of the savings? The interest rate is 10% per year.
3.38 Determine the present worth in year 0 of the cash
flows shown at an interest rate of 15% per year.
Year 1 2 3 4 5 6 7 8 9 10
Cash Flow, $ 90 90 90 85 80 75 70 65 60 55
3.39 The Pedernales Electric Cooperative estimates that
the present worth now of income from an investment
in renewable energy sources is $12,475,000.
There will be no income in years 1 and 2, but in
year 3 income will be $250,000, and thereafter it
will increase according to an arithmetic gradient
through year 15. What is the required gradient, if
the interest rate is 15% per year? Solve ( a ) by hand
and ( b ) using a spreadsheet.
3.40 Calculate the annual cost in years 1 through 9 of
the following series of disbursements. Use an interest
rate of 10% per year.
Year Disbursement, $ Year Disbursement, $
0 5000 5 7500
1 5500 6 8000
2 6000 7 8500
3 6500 8 9000
4 7000 9 9500
3.41 The cost associated with manufacturing highperformance
lubricants closely follows the cost of
crude oil. For the last 10 years, a small, independent
refiner had a cost of $3.4 million in
years 1 through 3, after which the cost increased by
3% per year through this year. Determine the current
equivalent worth (i.e., now ) of the manufacturing
cost, using an interest rate of 10% per year.
Show both ( a ) hand and ( b ) spreadsheet solutions.
3.42 Find the future worth in year 10 of $50,000 in
year 0 and amounts increasing by 15% per year
through year 10 at an interest rate of 10% per year.
3.43 The cost of tuition at public universities has been
steadily increasing for many years. One Midwestern
university pledged to keep the tuition constant for

Problems 91
4 years for all students who finished in the top 3%
of their class. One such student who liked research
planned to enroll at the university and continue
there until earning a PhD degree (a total time of
9 years). If the tuition for the first 4 years will be
$7200 per year and it increases by 5% per year for
the next 5 years, what is the present worth of the
tuition cost at an interest rate of 8% per year?
3.44 A private equity firm purchased a cable company
and assumed the company’s debt as part of the
transaction. The deal was structured such that the
private equity firm received $3 million immediately
after the deal was closed (in year 0) through
the sale of some assets. This year (year 1) income
was $3.36 million, and it is projected to increase
by 12% each year through expansion of the customer
base. What was the present worth in the year
of purchase of the income stream over a 10-year
period? The firm’s expected rate of return for any
purchase is 15% per year.
3.48 The pumping cost for delivering water from the
Ohio River to Wheeling Steel for cooling hotrolled
steel was $1.8 million for the first 3 years.
An effective energy conservation program resulted
in a reduced cost to $1.77 million in year 4,
$1.74 million in year 5, and amounts decreasing by
$30,000 each year through year 9. What is the
equivalent annual worth of the pumping costs over
the 9 years at an interest rate of 12% per year?
3.49 Income from the mining of mineral deposits usually
decreases as the resource becomes more difficult
to extract. Determine the future worth in
year 10 of a mineral lease that yielded income of
$14,000 in years 1 through 4 and then amounts
that decreased by 5% per year through year 10.
Use an interest rate of 18% per year. Show ( a ) hand
and ( b ) spreadsheet solutions.
3.50 For the cash flows shown in the diagram, determine
the future worth in year 8 at an interest rate of
10% per year.
i = 10% per year
3.45 Calculate the present worth in year 0 of a series of
cash flows that starts at $150,000 in year 0 and increases
by 10% per year through year 5. Assume
i 10% per year.
0
1
2 3 4 5 6 7 8
Year
Shifted Decreasing Gradients
3.46 For the cash flows shown, determine the value of
G such that the present worth in year 0 equals
$16,000 at an interest rate of 10% per year.
Year 0 1 2 3 4 5
Cash
0 8000 8000 8000 G 80002 G 80003 G
Flow, $
3.47 For these cash flows, find the equivalent annual
worth in years 1 through 7 at an interest rate of
10% per year.
Year Cash Flow, $ Year Cash Flow, $
0 850 4 650
1 800 5 600
2 750 6 550
3 700 7 500
$370
$320
$270
$220
$420
$470 $470
3.51 Income from the sale of application software (apps)
is usually constant for several years and then decreases
quite rapidly as the market gets close to
saturation. Income from one smart phone app was
$38,000 in years 1 through 3 and then decreased
geometrically by 15% per year through year 7. Determine
the equivalent annual income in years 1
through 7, using an interest rate of 10% per year.
3.52 Determine the future worth in year 10 of a cash
flow series that starts in year 0 at $100,000 and
decreases by 12% per year. Use an interest rate of
12% per year.

92 Chapter 3 Combining Factors and Spreadsheet Functions
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
3.53 A manufacturer of toilet flush valves wants to have
$1,900,000 available 3 years from now so that a
new product line can be initiated. If the company
plans to deposit money each year, starting now , the
equation that represents the deposit each year at
8% per year interest is:
(a) 1,900,000( AF ,8%,3)
(b) 1,900,000( AF ,8%,4)
(c) 1,900,000 1,900,000( AF ,8%,3)
(d) 1,900,000 1,900,000( AF ,8%,2)
3.54 The present worth in year 0 of a lease that requires
a payment of $9000 now and amounts increasing
by 5% per year through year 10 at 8% per year
interest is closest to:
( a ) $73,652
( b ) $79,939
( c ) $86,335
( d ) Over $87,000
3.55 For the diagram shown, the respective values of n
for the following equation are:
P 0 100( PA ,10%, n )( PF ,10%, n )
( a ) 6 and 1
( b ) 6 and 2
( c ) 7 and 1
( d ) 7 and 2
P 0 =?
0
1
i = 10% per year
2 3 4 5 6 7
Year
3.58 Cindy wants to deposit money for 4 consecutive
years starting 3 years from now so she can withdraw
$50,000 twelve years from now. Assume the interest
rate is 8% per year. The annual deposit is closest to:
( a ) $6990
( b ) $7670
( c ) $8530
( d ) $10,490
3.59 The net present worth in year 0 of the following
series of incomes and expenses at 8% per year is
closest to:
Years Income, $ Expenses, $
0 12,000 1000
1–6 700 100
7–11 900 200
( a ) $14,300 ( b ) $15,500
( c ) $16,100 ( d ) $16,500
3.60 For the cash flows shown, the equivalent annual
worth in periods 1 through 5 at an interest rate of
10% per year is closest to:
Annual Period Amount, $
0 1000
1 1000
2 1000
3 1000
4 1000
5 1500
( a ) $1120 ( b ) $1240
( c ) $1350 ( d ) $1490
$100 $100 $100 $100 $100 $100
3.56 Summit Metals is planning to expand its Wichita,
Kansas, manufacturing operation 5 years from now
at a cost of $10,000,000. If the company plans to deposit
money into an account each year for 4 years
beginning 2 years from now (first deposit is in year 2)
to pay for the expansion, the equation that represents
the amount of the deposit at 9% per year interest is:
(a) A 10,000,000( AF ,9%,5)
(b) A 10,000,000( AF ,9%,4)
(c) A 10,000,000( AP ,9%,4)
(d) A 10,000,000( AF ,9%,4)( PF ,9%,1)
3.57 The amount of money a person must deposit
3 years from now in order to withdraw $10,000 per
year for 10 years beginning 15 years from now at
an interest rate of 10% per year is closest to:
( a ) $15,500 ( b ) $17,200
( c ) $19,300 ( d ) $21,500
3.61 For the cash flows shown, the value of X that will
make the present worth in year 0 equal to $5000 at
an interest rate of 10% per year is closest to:
Year 0 1 2 3 4 5 6 7 8 9
Cash 200 300 400 500 600 700 800 900 1000 X
Flow, $
( a ) $2895 ( b ) $3125
( c ) $3305 ( d ) $3765
3.62 In order to have cash available for unforeseen
emergencies, Baring Systems, a military contractor,
wants to have $2,000,000 in a contingency
fund 4 years from now. The amount the company
must deposit each year in years 0 through 4 at an
interest rate of 10% per year is closest to:
( a ) $420,100 ( b ) $327,600
( c ) $284,600 ( d ) $206,900

Case Study 93
CASE STUDY
PRESERVING LAND FOR PUBLIC USE
Background and Information
The Trust for Public Land (TPL) is a national organization
that purchases and oversees the improvement of large land
sites for government agencies at all levels. Its mission is to
ensure the preservation of the natural resources, while providing
necessary, but minimal, development for recreational
use by the public. All TPL projects are evaluated at 7% per
year, and TPL reserve funds earn 7% per year.
A southern U.S. state, which has long-term groundwater
problems, has asked the TPL to manage the purchase of
10,000 acres of aquifer recharge land and the development of
three parks of different use types on the land. The 10,000
acres will be acquired in increments over the next 5 years
with $4 million expended immediately on purchases. Total
annual purchase amounts are expected to decrease 25% each
year through year 5 and then cease for this particular project.
A city with 1.5 million citizens immediately to the southeast
of this acreage relies heavily on the aquifer’s water. Its
citizens passed a bond issue last year, and the city government
now has available $3 million for the purchase of land.
The bond interest rate is an effective 7% per year.
The engineers working on the park plan intend to complete
all the development over a 3-year period starting in year
4, when the amount budgeted is $550,000. Increases in construction
costs are expected to be $100,000 each year through
year 6.
At a recent meeting, the following agreements were made:
• Purchase the initial land increment now. Use the bond
issue funds to assist with this purchase. Take the remaining
amount from TPL reserves.
• Raise the remaining project funds over the next 2 years in
equal annual amounts.
• Evaluate a financing alternative (suggested informally by
one individual at the meeting) in which the TPL provides
all funds, except the $3 million available now, until the
parks development is initiated in year 4.
Case Study Exercises
1. For each of the 2 years, what is the equivalent annual
amount necessary to supply the remaining project
funds?
2. If the TPL did agree to fund all costs except the $3 million
bond proceeds now available, determine the equivalent
annual amount that must be raised in years 4
through 6 to supply all remaining project funds. Assume
the TPL will not charge any extra interest over the 7% to
the state or city on the borrowed funds.
3. Review the TPL website (www.tpl.org). Identify some
economic and noneconomic factors that you believe
must be considered when the TPL is deciding to purchase
land to protect it from real estate development.

CHAPTER 4
Nominal and
Effective
Interest Rates
LEARNING OUTCOMES
Purpose: Make computations for interest rates and cash flows that are on a time basis other than a year.
SECTION TOPIC LEARNING OUTCOME
4.1 Statements • Understand interest rate statements that include
nominal and effective rates.
4.2 Effective annual rate • Derive and use the formula for an effective
annual interest rate.
4.3 Effective rate • Determine the effective interest rate for any
stated time period.
4.4 Payment period and
compounding period
• Determine the payment period (PP) and
compounding period (CP) for equivalence
computations.
4.5 Single cash flows with PP CP • Perform equivalence calculations for singleamount
cash flows and PP CP.
4.6 Series cash flows with PP CP • Perform equivalence calculations for series and
gradient cash flows and PP CP.
4.7 Single amounts and series with
PP CP
• Perform equivalence calculations for cash flows
with PP CP.
4.8 Continuous compounding • Derive and use the effective interest rate formula
for interest rates that are compounded
continuously.
4.9 Varying rates • Perform equivalency calculations for interest rates
that vary from one time period to another.

I
n all engineering economy relations developed thus far, the interest rate has been
a constant, annual value. For a substantial percentage of the projects evaluated by
professional engineers in practice, the interest rate is compounded more frequently
than once a year; frequencies such as semiannually, quarterly, and monthly are common. In fact,
weekly, daily, and even continuous compounding may be experienced in some project evaluations.
Also, in our own personal lives, many of our financial considerations—loans of all types
(home mortgages, credit cards, automobiles, boats), checking and savings accounts, investments,
stock option plans, etc.—have interest rates compounded for a time period shorter than
1 year. This requires the introduction of two new terms— nominal and effective interest rates.
This chapter explains how to understand and use nominal and effective interest rates
in engineering practice and in daily life situations. Equivalence calculations for any compounding
frequency in combination with any cash flow frequency are presented.
PE
The Credit Card Offer Case: Today, Dave
received a special offer of a new credit
card from Chase Bank linked with the
major airline that he flies frequently.
It offers a generous bonus package
for signing up by a specific date about
60 days from now. The bonus package
includes extra airline points (once the
first purchase is made), priority airport
check-in services (for 1 year), several
free checked-bag allowances (for up
to 10 check-ins), extra frequent-flyer
points on the airline, access to airline
lounges (provided he uses the card
on a set time basis), plus several other
rewards (rental car discounts, cruise trip
amenities, and floral order discounts).
The annual fee of $85 for membership
does not start until the second year,
and balance transfers from other credit
cards have a low transfer fee, provided
they are made at the time of initial
membership.
Dave has a credit card currently with
a bank that he is planning to leave due
to its poor customer service and high
monthly fees. If he enrolls, he will transfer
the $1000 balance on the current
card to the new Chase Bank card.
In the page that accompanies the
offer letter, “pricing information” is
included. This includes interest rates,
interest charges, and fees. A summary of
several of these rates and fees follows.
APR (annual percentage rate) for purchases
and balance transfers*
APR for cash and overdraft advances*
Penalty APR for late minimum payment,
exceeding credit limit, and
returned unpaid payments*†
14.24% per year (sum of the current U.S.
Government prime rate of 3.25% and
10.99%, which is the APR added to determine
the balance transfer APR for Chase Bank)
19.24% per year
29.99% per year (maximum penalty APR)
Fees are listed as follows:
Annual membership
$85; free the first year
Balance transfers
$5 or 3% of each transfer, whichever is greater
Cash advances
$10 or 3% of each advance, whichever is greater
Late payment $39 each occurrence, if balance exceeds $250
Over the credit limit
$39 each occurrence
Returned check or payment $39 each occurrence
*All APR rates are variable, based on a current 3.25% prime rate with 10.99% added to
determine purchasebalance transfer APR; with 15.99% added to determine cashoverdraft
APR; and with 26.99% added to determine penalty APR.
†The penalty APR applies indefinitely to future transactions. If no minimum payment is received
within 60 days, the penalty APR applies to all outstanding balances and all future transactions
on the account.
(Continued)

96 Chapter 4 Nominal and Effective Interest Rates
This case is used in the following topics
(and sections) of this chapter:
Nominal and effective interest rate
statements (4.1)
Effective annual interest rates (4.2)
Equivalence relations: Series with
PP CP (4.6)
4.1 Nominal and Effective Interest Rate Statements
In Chapter 1, we learned that the primary difference between simple interest and compound
interest is that compound interest includes interest on the interest earned in the previous period,
while simple interest does not. Here we discuss nominal and effective interest rates, which
have the same basic relationship. The difference here is that the concepts of nominal and effective
must be used when interest is compounded more than once each year. For example, if an
interest rate is expressed as 1% per month, the terms nominal and effective interest rates must
be considered.
To understand and correctly handle effective interest rates is important in engineering practice
as well as for individual finances. The interest amounts for loans, mortgages, bonds, and stocks
are commonly based upon interest rates compounded more frequently than annually. The engineering
economy study must account for these effects. In our own personal finances, we manage
most cash disbursements and receipts on a nonannual time basis. Again, the effect of compounding
more frequently than once per year is present. First, consider a nominal interest rate.
Nominal interest rate r
A nominal interest rate r is an interest rate that does not account for compounding. By definition,
r interest rate per time period number of periods [4.1]
A nominal rate may be calculated for any time period longer than the time period stated by using
Equation [4.1]. For example, the interest rate of 1.5% per month is the same as each of the following
nominal rates.
Time Period
Nominal Rate by
Equation [4.1]
What This Is
24 months 1.5 24 36% Nominal rate per 2 years
12 months 1.5 12 18% Nominal rate per 1 year
6 months 1.5 6 9% Nominal rate per 6 months
3 months 1.5 3 4.5% Nominal rate per 3 months
Note that none of these rates mention anything about compounding of interest; they are all of the
form “ r % per time period.” These nominal rates are calculated in the same way that simple rates
are calculated using Equation [1.7], that is, interest rate times number of periods.
After the nominal rate has been calculated, the compounding period (CP) must be included
in the interest rate statement. As an illustration, again consider the nominal rate of 1.5%
per month. If we define the CP as 1 month, the nominal rate statement is 18% per year, compounded
monthly, or 4.5% per quarter, compounded monthly . Now we can consider an effective
interest rate.
Effective interest rate i
An effective interest rate i is a rate wherein the compounding of interest is taken into
account. Effective rates are commonly expressed on an annual basis as an effective annual
rate; however, any time basis may be used.
The most common form of interest rate statement when compounding occurs over time periods
shorter than 1 year is “% per time period, compounded CP-ly,” for example, 10% per year, compounded
monthly, or 12% per year, compounded weekly. An effective rate may not always
include the compounding period in the statement. If the CP is not mentioned, it is understood to

4.1 Nominal and Effective Interest Rate Statements 97
be the same as the time period mentioned with the interest rate. For example, an interest rate of
“1.5% per month” means that interest is compounded each month; that is, CP is 1 month. An
equivalent effective rate statement, therefore, is 1.5% per month, compounded monthly.
All of the following are effective interest rate statements because either they state they are
effective or the compounding period is not mentioned. In the latter case, the CP is the same as
the time period of the interest rate.
Statement CP What This Is
i 10% per year CP not stated; CP year Effective rate per year
i effective 10% per year, CP stated; CP month Effective rate per year
compounded monthly
i 1
1_ % per month
2
CP not stated; CP month Effective rate per month
i effective 1 1_ % per month,
2
compounded monthly
CP stated; CP month Effective rate per month; terms effective
and compounded monthly are redundant
i effective 3% per quarter,
compounded daily
CP stated; CP day Effective rate per quarter
All nominal interest rates can be converted to effective rates. The formula to do this is discussed
in the next section.
All interest formulas, factors, tabulated values, and spreadsheet functions must use an effective
interest rate to properly account for the time value of money.
The term APR (Annual Percentage Rate) is often stated as the annual interest rate for credit
cards, loans, and house mortgages. This is the same as the nominal rate . An APR of 15% is the
same as a nominal 15% per year or a nominal 1.25% on a monthly basis. Also the term APY
(Annual Percentage Yield) is a commonly stated annual rate of return for investments, certificates
of deposit, and saving accounts. This is the same as an effective rate . The names are different,
but the interpretations are identical. As we will learn in the following sections, the effective
rate is always greater than or equal to the nominal rate, and similarly APY APR.
Based on these descriptions, there are always three time-based units associated with an interest
rate statement.
Interest period ( t )—The period of time over which the interest is expressed. This is the t in
the statement of r % per time period t , for example, 1% per month. The time unit of 1 year is
by far the most common. It is assumed when not stated otherwise.
Compounding period (CP)—The shortest time unit over which interest is charged or earned.
This is defined by the compounding term in the interest rate statement, for example, 8% per
year, compounded monthly. If CP is not stated, it is assumed to be the same as the interest
period.
Compounding frequency (m)—The number of times that compounding occurs within the
interest period t. If the compounding period CP and the time period t are the same, the compounding
frequency is 1, for example, 1% per month, compounded monthly.
Consider the (nominal) rate of 8% per year, compounded monthly. It has an interest period t of
1 year, a compounding period CP of 1 month, and a compounding frequency m of 12 times per
year. A rate of 6% per year, compounded weekly, has t 1 year, CP 1 week, and m 52, based
on the standard of 52 weeks per year.
In previous chapters, all interest rates had t and CP values of 1 year, so the compounding frequency
was always m 1. This made them all effective rates, because the interest period and
compounding period were the same. Now, it will be necessary to express a nominal rate as an
effective rate on the same time base as the compounding period.
An effective rate can be determined from a nominal rate by using the relation
Effective rate per CP
r % per time period t
—————————————
m compounding periods per t r —
m [4.2]
As an illustration, assume r 9% per year, compounded monthly; then m 12. Equation
[4.2] is used to obtain the effective rate of 9%12 0.75% per month, compounded monthly.

98 Chapter 4 Nominal and Effective Interest Rates
Note that changing the interest period t does not alter the compounding period, which is 1 month
in this illustration. Therefore, r 9% per year, compounded monthly, and r 4.5% per 6 months,
compounded monthly, are two expression of the same interest rate.
EXAMPLE 4.1
Three different bank loan rates for electric generation equipment are listed below. Determine
the effective rate on the basis of the compounding period for each rate.
(a) 9% per year, compounded quarterly.
(b) 9% per year, compounded monthly.
(c) 4.5% per 6 months, compounded weekly.
Solution
Apply Equation [4.2] to determine the effective rate per CP for different compounding periods.
The graphic in Figure 4–1 indicates the effective rate per CP and how the interest rate is distributed
over time.
Effective
Nominal Compounding Compounding Rate per
r% per t Period (CP) Frequency (m) CP ( —
m r )
Distribution over Time Period t
(a) 9% per Quarter 4 2.25%
year
2.25%
2.25% 2.25% 2.25%
1 2 3 4
Quarter
(b) 9% per Month 12 0.75%
year
.75%
.75%
.75%
.75%
.75%
.75%
.75%
.75%
.75%
.75%
.75%
.75%
1 2 3 4 5 6 7 8 9 10 11 12
0.173%
Month
(c) 4.5% per Week 26 0.173%
6 months
1 12 14 16
26
Week
Figure 4–1
Relations between interest period t, compounding period CP, and effective interest rate per CP.
Sometimes it is not obvious whether a stated rate is a nominal or an effective rate. Basically
there are three ways to express interest rates, as detailed in Table 4–1. The right column includes
a statement about the effective rate. For the first format, a nominal interest rate is given and the
compounding period is stated. The effective rate must be calculated (discussed in the next sections).
In the second format, the stated rate is identified as effective (or APY could also be used),
so the rate is used directly in computations.
In the third format, no compounding period is identified, for example, 8% per year. This rate
is effective over a compounding period equal to the stated interest period of 1 year in this case.
The effective rate for any other time period must be calculated.
TABLE 4–1
Various Ways to Express Nominal and Effective Interest Rates
Format of Rate Statement Example of Statement What about the Effective Rate?
Nominal rate stated,
compounding period stated
Effective rate stated
Interest rate stated, no
compounding period stated
8% per year, compounded
quarterly
Effective 8.243% per year,
compounded quarterly
Find effective rate for any time
period (next two sections)
Use effective rate of 8.243% per year
directly for annual cash flows
8% per year Rate is effective for CP equal to stated
interest period of 1 year; find effective
rate for all other time periods

4.2 Effective Annual Interest Rates 99
EXAMPLE 4.2 The Credit Card Offer Case
As described in the introduction to this case, Dave has been offered what is described as a credit
card deal that should not be refused—at least that is what the Chase Bank offer letter implies. The
balance transfer APR interest rate of 14.24% is an annual rate, with no compounding period mentioned.
Therefore, it follows the format of the third entry in Table 4–1, that is, interest rate stated,
no CP stated. Therefore, we should conclude that the CP is 1 year, the same as the annual interest
period of the APR. However, as Dave and we all know, credit card payments are required monthly.
(a) First, determine the effective interest rates for compounding periods of 1 year and
1 month so Dave knows some effective rates he might be paying when he transfers the
$1000 balance from his current card.
(b) Second, assume that immediately after he accepts the card and completes the $1000 transfer,
Dave gets a bill that is due 1 month later. What is the amount of the total balance he owes?
PE
Now, Dave looks a little closer at the fine print of the “pricing information” sheet and discovers
a small-print statement that Chase Bank uses the daily balance method (including new transactions)
to determine the balance used to calculate the interest due at payment time.
(c) We will reserve the implication of this new finding until later, but for now help Dave by
determining the effective daily interest rate that may be used to calculate interest due at
the end of 1 month, provided the CP is 1 day.
Solution
(a) The interest period is 1 year. Apply Equation [4.2] for both CP values of 1 year ( m 1
compounding period per year) and 1 month ( m 12 compounding periods per year).
CP of year: Effective rate per year 14.241 14.24%
CP of month: Effective rate per month 14.2412 1.187%
(b) The interest will be at the monthly effective rate, plus the balance transfer fee of 3%.
Amount owed after 1 month 1000 1000(0.01187) 0.03(1000)
1000 11.87 30
$1041.87
Including the $30 fee, this represents an interest rate of (41.871000)(100%) 4.187%
for only the 1-month period.
(c) Again apply Equation [4.2], now with m 365 compounding periods per year.
CP of day: Effective rate per day 14.24365 0.039%
4.2 Effective Annual Interest Rates
In this section, effective annual interest rates are calculated. Therefore, the year is used as the interest
period t , and the compounding period CP can be any time unit less than 1 year. For example,
we will learn that a nominal 18% per year, compounded quarterly is the same as an effective rate of
19.252% per year.
The symbols used for nominal and effective interest rates are
r nominal interest rate per year
CP time period for each compounding
m number of compounding periods per year
i effective interest rate per compounding period rm
i a effective interest rate per year
The relation i rm is exactly the same as Equation [4.2].

100 Chapter 4 Nominal and Effective Interest Rates
Figure 4–2
Future worth calculation
at a rate i , compounded m
times in a year.
P(1 + i) m = P(1 + i a )
P(1 + i) m –1
P(1 + i) m –2
P(1 + i) 3
P(1 + i) 2 Future worth amounts
P(1 + i)
P
1 2 3 m – 2 m – 1 m Number of CPs
i i i i i i
Effective i per
compounding period
1 2 3 m – 2 m – 1 m
Compounding period
As mentioned earlier, treatment for nominal and effective interest rates parallels that of
simple and compound interest. Like compound interest, an effective interest rate at any point
during the year includes (compounds) the interest rate for all previous compounding periods
during the year. Therefore, the derivation of an effective interest rate formula directly parallels
the logic used to develop the future worth relation F P (1 i ) n . We set P $1 for
simplification.
The future worth F at the end of 1 year is the principal P plus the interest P ( i ) through the
year. Since interest may be compounded several times during the year, use the effective annual
rate symbol i a to write the relation for F with P $1.
F P Pi a 1 (1 i a )
Now consider Figure 4–2. The effective rate i per CP must be compounded through all m periods
to obtain the total effect of compounding by the end of the year. This means that F can also be
written as
F 1 (1 i )
m
Equate the two expressions for F and solve for i a . The effective annual interest rate formula
for i a is
i a (1 i )
m
1 [4.3]
Equation [4.3] calculates the effective annual interest rate i a for any number of compounding
periods per year when i is the rate for one compounding period.
If the effective annual rate i a and compounding frequency m are known, Equation [4.3] can be
solved for i to determine the effective interest rate per compounding period.
i (1 i a ) 1 m
1 [4.4]
As an illustration, Table 4–2 utilizes the nominal rate of 18% per year for different compounding
periods (year to week) to determine the effective annual interest rate. In each case, the effective
rate i per CP is applied m times during the year. Table 4–3 summarizes the effective annual
rate for frequently quoted nominal rates using Equation [4.3]. A standard of 52 weeks and
365 days per year is used throughout. The values in the continuous-compounding column are
discussed in Section 4.8.

TABLE 4–2 Effective Annual Interest Rates Using Equation [4.3]
r 18% per year, compounded CP-ly
Compounding
Period, CP
Times
Compounded
per Year, m
Rate per
Compound
Period, i%
Distribution of i over the Year of
Compounding Periods
Effective Annual
Rate, i a (1 i ) m 1
Year 1 18 18%
(1.18) 1 1 18%
1
6 months 2 9 9%
9%
(1.09) 2 1 18.81%
1 2
Quarter 4 4.5 4.5% 4.5% 4.5% 4.5% (1.045) 4 1 19.252%
1 2 3 4
1.5% in each
Month 12 1.5 (1.015) 12 1 19.562%
1
2 3 4 5 6 7 8 9 10 11 12
0.34615% in each
Week 52 0.34615 (1.0034615) 52 1 19.684%
1 2 3 24 26 28 50 52
101

102 Chapter 4 Nominal and Effective Interest Rates
TABLE 4–3
Nominal
Rate r %
Effective Annual Interest Rates for Selected Nominal Rates
Semiannually
(m 2)
Quarterly
(m 4)
Monthly
(m 12)
Weekly
(m 52)
Daily
(m 365)
Continuously
(m ; e r 1)
0.25 0.250 0.250 0.250 0.250 0.250 0.250
0.50 0.501 0.501 0.501 0.501 0.501 0.501
1.00 1.003 1.004 1.005 1.005 1.005 1.005
1.50 1.506 1.508 1.510 1.511 1.511 1.511
2 2.010 2.015 2.018 2.020 2.020 2.020
3 3.023 3.034 3.042 3.044 3.045 3.046
4 4.040 4.060 4.074 4.079 4.081 4.081
5 5.063 5.095 5.116 5.124 5.126 5.127
6 6.090 6.136 6.168 6.180 6.180 6.184
7 7.123 7.186 7.229 7.246 7.247 7.251
8 8.160 8.243 8.300 8.322 8.328 8.329
9 9.203 9.308 9.381 9.409 9.417 9.417
10 10.250 10.381 10.471 10.506 10.516 10.517
12 12.360 12.551 12.683 12.734 12.745 12.750
15 15.563 15.865 16.076 16.158 16.177 16.183
18 18.810 19.252 19.562 19.684 19.714 19.722
20 21.000 21.551 21.939 22.093 22.132 22.140
25 26.563 27.443 28.073 28.325 28.390 28.403
30 32.250 33.547 34.489 34.869 34.968 34.986
40 44.000 46.410 48.213 48.954 49.150 49.182
50 56.250 60.181 63.209 64.479 64.816 64.872
EXAMPLE 4.3
Janice is an engineer with Southwest Airlines. She purchased Southwest stock for $6.90 per share
and sold it exactly 1 year later for $13.14 per share. She was very pleased with her investment
earnings. Help Janice understand exactly what she earned in terms of ( a ) effective annual rate and
( b ) effective rate for quarterly compounding, and for monthly compounding. Neglect any commission
fees for purchase and selling of stock and any quarterly dividends paid to stockholders.
Solution
(a) The effective annual rate of return i a has a compounding period of 1 year, since the
stock purchase and sales dates are exactly 1 year apart. Based on the purchase price of
$6.90 per share and using the definition of interest rate in Equation [1.2],
amount of increase per 1 year
i a ———————————— 100% —— 6.24 100% 90.43% per year
original price
6.90
(b) For the effective annual rates of 90.43% per year, compounded quarterly, and 90.43%,
compounded monthly, apply Equation [4.4] to find corresponding effective rates on the
basis of each compounding period.
Quarter: m 4 times per year i (1.9043) 14 1 1.17472 1 0.17472
This is 17.472% per quarter, compounded quarterly.
Month: m 12 times per year i (1.9043) 112 1 1.05514 1 0.05514
This is 5.514% per month, compounded monthly.
Comment
Note that these quarterly and monthly rates are less than the effective annual rate divided by
the number of quarters or months per year. In the case of months, this would be 90.43%12
7.54% per month. This computation is incorrect because it neglects the fact that compounding
takes place 12 times during the year to result in the effective annual rate of 90.43%.
The spreadsheet function that displays the result of Equation [4.3], that is, the effective annual
rate i a , is the EFFECT function. The format is

4.2 Effective Annual Interest Rates 103
EFFECT(nominal_rate_per_year, compounding_frequency)
EFFECT( r %, m ) [4.5]
Note that the rate entered in the EFFECT function is the nominal annual rate r% per year , not
the effective rate i% per compounding period. The function automatically finds i for use in Equation
[4.3]. As an example, assume the nominal annual rate is r 5.25% per year, compounded
quarterly, and you want to find the effective annual rate i a . The correct input is EFFECT(5.25%,4)
to display i a 5.354% per year. This is the spreadsheet equivalent of Equation [4.3] with
i 5.254 1.3125% per quarter with m 4.
i a (1 0.013125) 4 1 0.05354 (5.354%)
The thing to remember about using the EFFECT function is that the nominal rate r entered
must be expressed over the same period of time as that of the effective rate required, which is
1 year here.
The NOMINAL spreadsheet function finds the nominal annual rate r . The format is
NOMINAL(effective_rate, compounding_frequency_per_year)
NOMINAL( i a %, m ) [4.6]
This function is designed to display only nominal annual rates. Accordingly, the m entered must
be the number of times interest is compounded per year. For example, if the effective annual rate
is 10.381% per year, compounded quarterly, and the nominal annual rate is requested, the function
is NOMINAL(10.381%,4) to display r 10% per year, compounded quarterly. The
nominal rates for shorter time periods than 1 year are determined by using Equation [4.1]. For
example, the quarterly rate is 10%4 2.5%.
The things to remember when using the NOMINAL function are that the answer is always a
nominal annual rate, the rate entered must be an effective annual rate, and the m must equal the
number of times interest is compounded annually.
EXAMPLE 4.4 The Credit Card Offer Case
In our Progressive Example, Dave is planning to accept the offer for a Chase Bank credit card
that carries an APR (nominal rate) of 14.24% per year, or 1.187% per month. He will transfer
a balance of $1000 and plans to pay it and the transfer fee of $30, due at the end of the first
month. Let’s assume that Dave makes the transfer, and only days later his employer has a
1-year assignment for him in the country of the Cameroon in northwestern Africa. Dave accepts
the employment offer, and in his hurried, excited departure, he forgets to send the credit
card service company a change of address. Since he is now out of mail touch, he does not pay
his monthly balance due, which we calculated in Example 4.2 to be $1041.87.
(a) If this situation continues for a total of 12 months, determine the total due after 12 months
and the effective annual rate of interest Dave has accumulated. Remember, the fine print on
the card’s interest and fee information states a penalty APR of 29.99% per year after one late
payment of the minimum payment amount, plus a late payment fee of $39 per occurrence.
(b) If there were no penalty APR and no late-payment fee, what effective annual interest rate
would be charged for this year? Compare this rate with the answer in part ( a ).
PE
Solution
(a) Because Dave did not pay the first month’s amount, the new balance of $1041.87 and all
future monthly balances will accumulate interest at the higher monthly rate of
29.99%12 2.499% per month
Additionally, the $39 late-payment fee will be added each month, starting with the second
month, and interest will be charged on these fees also each month thereafter. The first
3 months and last 2 months are detailed below. Figure 4–3 details the interest and fees for
all 12 months.

104 Chapter 4 Nominal and Effective Interest Rates
SUM(B6:D6)
J4/12
K4/12
Interest rate per month
Month 1: 1.187%
Months 2-12: penalty 2.499%
Month 1: transfer fee
Months 2-12: late-payment fee
Figure 4–3
Monthly amounts due for a credit card, Example 4.4.
Month 1: 1000 1000(0.01187) 30 $1041.87
Month 2: 1041.87 1041.87(0.02499) 39 $1106.91
Month 3: 1106.91 1106.91(0.02499) 39 $1173.57
. . .
Month 11: 1689.25 1689.25(0.02499) 39 $1770.46
Month 12: 1770.46 1770.46(0.02499) 39 $1853.71
The effective monthly rate is determined by using the FP factor to find the i value at
which $1000 now is equivalent to $1853.71 after 12 periods.
1853.71 1000( FP, i ,12) 1000(1 i ) 12
1 i (1.85371 ) 112 1.05278
i 5.278% per month
Since the compounding period is 1 month, use Equation [4.3] to determine the effective
annual rate of 85.375% per year, compounded monthly.
i a (1 i ) m 1 (1.05278 ) 12 1
0.85375 (85.375%)
(b) If there were no penalty fees for late payments and the nominal annual rate of 14.24% (or
1.187% per month) were applied throughout the 12 months, the effective annual rate
would be 15.207% per year, compounded monthly. By Equation [4.3], with a small
rounding error included,
i a (1 i ) m 1 (1.01187) 12 1 0.15207
First, Dave will not pay at the stated rate of 14.24%, because this is the APR (nominal
rate), not the APY (effective rate) of 15.207%. Second, and much more important, is the
huge difference made by (1) the increase in rate to an APR of 29.99% and (2) the monthly
fees of $39 for not making a payment. These large fees become part of the credit balance
and accumulate interest at the penalty rate of 29.99% per year. The result is an effective
annual rate jump from 15.207% to 85.375% per year, compounded monthly.
Comment
This is but one illustration of why the best advice to an individual or company in debt is to
spend down the debt. The quoted APR by credit card, loan, and mortgage institutions can be
quite deceiving; plus, the addition of penalty fees increases the effective rate very rapidly.

4.3 Effective Interest Rates for Any Time Period 105
When Equation [4.3] is applied to find i a the result is usually not an integer. Therefore, the
engineering economy factor cannot be obtained directly from the interest factor tables. There are
alternative ways to find the factor value.
• Use the factor formula with the ia rate substituted for i .
• Use the spreadsheet function with i a (as discussed in Section 2.4).
• Linearly interpolate between two tabulated rates (as discussed in Section 2.4).
4.3 Effective Interest Rates for Any Time Period
Equation [4.3] in Section 4.2 calculates an effective interest rate per year from any effective rate
over a shorter time period. We can generalize this equation to determine the effective interest
rate for any time period (shorter or longer than 1 year).
Effective i per time period ( 1 r —
m ) m 1 [4.7]
where i effective rate for specified time period (say, semiannual)
r nominal interest rate for same time period (semiannual)
m number of times interest is compounded per stated time period (times per
6 months)
The term rm is always the effective interest rate over a compounding period CP, and m is
always the number of times that interest is compounded per the time period on the left of the
equals sign in Equation [4.7]. Instead of i a , this general expression uses i as the symbol for the
effective interest rate, which conforms to the use of i throughout the remainder of this text.
Examples 4.5 and 4.6 illustrate the use of this equation.
EXAMPLE 4.5
Tesla Motors manufactures high-performance battery electric vehicles. An engineer is on a
Tesla committee to evaluate bids for new-generation coordinate-measuring machinery to be
directly linked to the automated manufacturing of high-precision vehicle components. Three
bids include the interest rates that vendors will charge on unpaid balances. To get a clear understanding
of finance costs, Tesla management asked the engineer to determine the effective
semiannual and annual interest rates for each bid. The bids are as follows:
Bid 1:
Bid 2:
Bid 3:
9% per year, compounded quarterly
3% per quarter, compounded quarterly
8.8% per year, compounded monthly
(a) Determine the effective rate for each bid on the basis of semiannual periods.
(b) What are the effective annual rates? These are to be a part of the final bid selection.
(c) Which bid has the lowest effective annual rate?
Solution
(a) Convert the nominal rates to a semiannual basis, determine m, then use Equation [4.7] to
calculate the effective semiannual interest rate i. For bid 1,
r 9% per year 4.5% per 6 months
m 2 quarters per 6 months
Effective i% per 6 months ( 1 0.045 ———
2 ) 2 1 1.0455 1 4.55%
Table 4–4 (left section) summarizes the effective semiannual rates for all three bids.

106 Chapter 4 Nominal and Effective Interest Rates
TABLE 4–4 Effective Semiannual and Annual Interest Rates for Three Bid Rates, Example 4.5
Bid
Nominal r per
6 Months, %
Semiannual Rates
CP per
6 Months, m
Equation [4.7],
Effective i, %
Nominal r
per Year, %
Annual Rates
CP per
Year, m
Equation [4.7],
Effective i, %
1 4.5 2 4.55 9 4 9.31
2 6.0 2 6.09 12 4 12.55
3 4.4 6 4.48 8.8 12 9.16
(b) For the effective annual rate, the time basis in Equation [4.7] is 1 year. For bid 1,
r 9% per year
m 4 quarters per year
Effective i% per year ( 1 0.09 ——
4 ) 4 − 1 1.0931 1 9.31%
The right section of Table 4–4 includes a summary of the effective annual rates.
(c) Bid 3 includes the lowest effective annual rate of 9.16%, which is equivalent to an effective
semiannual rate of 4.48% when interest is compounded monthly.
EXAMPLE 4.6
A dot-com company plans to place money in a new venture capital fund that currently returns
18% per year, compounded daily. What effective rate is this ( a ) yearly and ( b ) semiannually?
Solution
(a) Use Equation [4.7], with r 0.18 and m 365.
Effective i % per year ( 1 —— 0.18
365 ) 365 1 19.716%
(b) Here r 0.09 per 6 months and m 182 days.
Effective i % per 6 months ( 1 —— 0.09
182 ) 182 1 9.415%
4.4 Equivalence Relations: Payment Period
and Compounding Period
Now that the procedures and formulas for determining effective interest rates with consideration
of the compounding period are developed, it is necessary to consider the payment period .
The payment period (PP) is the length of time between cash flows (inflows or outflows). It is
common that the lengths of the payment period and the compounding period (CP) do not coincide.
It is important to determine if PP CP, PP CP, or PP CP.
If a company deposits money each month into an account that earns at the nominal rate of 8%
per year, compounded semiannually, the cash flow deposits define a payment period of 1 month
and the nominal interest rate defines a compounding period of 6 months. These time periods are
shown in Figure 4–4. Similarly, if a person deposits a bonus check once a year into an account
that compounds interest quarterly, PP 1 year and CP 3 months.
r = nominal 8% per year, compounded semiannually
CP
6 months
CP
6 months
Figure 4–4
One-year cash flow diagram
for a monthly payment
period (PP) and
semiannual compounding
period (CP).
0 1 2 3 4 5 6 7 8 9 10 11 12
PP
1 month
Months

4.5 Equivalence Relations: Single Amounts with PP CP 107
TABLE 4–5
Length
of Time
Section References for Equivalence Calculations Based on
Payment Period and Compounding Period Comparison
Involves
Single Amounts
(P and F Only)
PP CP Section 4.5 Section 4.6
PP > CP Section 4.5 Section 4.6
PP < CP Section 4.7 Section 4.7
Involves Uniform Series
or Gradient Series
(A, G, or g)
As we learned earlier, to correctly perform equivalence calculations, an effective interest rate
is needed in the factors and spreadsheet functions. Therefore, it is essential that the time periods
of the interest rate and the payment period be on the same time basis. The next three sections (4.5
to 4.7) describe procedures to determine correct i and n values for engineering economy factors
and spreadsheet functions. First, compare the length of PP and CP, then identify the cash flows as
only single amounts ( P and F ) or as a series ( A , G , or g ). Table 4–5 provides the section reference.
The section references are the same when PP CP and PP > CP, because the procedures to
determine i and n are the same, as discussed in Sections 4.5 and 4.6.
A general principle to remember throughout these equivalence computations is that when
cash actually flows, it is necessary to account for the time value of money. For example, assume
that cash flows occur every 6 months and that interest is compounded quarterly. After 3
months there is no cash flow and no need to determine the effect of quarterly compounding.
However, at the 6-month time point, it is necessary to consider the interest accrued during the
previous two quarters.
4.5 Equivalence Relations: Single Amounts with PP CP
With only P and F estimates defined, the payment period is not specifically identified. In virtually
all situations, PP will be equal to or greater than CP. The length of the PP is defined by the interest
period in the stated interest rate. If the rate is 8% per year, for example, PP CP 1 year. However,
if the rate is 10% per year, compounded quarterly, then PP is 1 year, CP is 1 quarter or
3 months, and PP > CP. The procedures to perform equivalence computations are the same for
both situations, as explained here.
When only single-amount cash flows are involved, there are two equally correct ways to determine
i and n for P F and F P factors. Method 1 is easier to apply, because the interest tables
in the back of the text can usually provide the factor value. Method 2 likely requires a factor
formula calculation, because the resulting effective interest rate is not an integer. For spreadsheets,
either method is acceptable; however, method 1 is usually easier.
Method 1: Determine the effective interest rate over the compounding period CP , and set n
equal to the number of compounding periods between P and F . The relations to calculate P and
F are
P F ( P F , effective i % per CP, total number of periods n ) [4.8]
F P ( F P , effective i % per CP, total number of periods n ) [4.9]
For example, assume that the stated credit card rate is nominal 15% per year, compounded
monthly. Here CP is 1 month. To find P or F over a 2-year span, calculate the effective monthly
rate of 15%12 1.25% and the total months of 2(12) 24. Then 1.25% and 24 are used in the
PF and FP factors.
Any time period can be used to determine the effective interest rate; however, the interest rate
that is associated with the CP is typically the best because it is usually a whole number. Therefore,
the factor tables in the back of the text can be used.
Method 2: Determine the effective interest rate for the time period t of the nominal rate, and
set n equal to the total number of periods, using this same time period.

108 Chapter 4 Nominal and Effective Interest Rates
The P and F relations are the same as in Equations [4.8] and [4.9] with the term effective i% per t
substituted for the interest rate. For a credit card rate of 15% per year, compounded monthly, the
time period t is 1 year. The effective rate over 1 year and the n values are
Effective i % per year ( 1 0.15 ——
12 ) 12 1 16.076%
n 2 years
The P F factor is the same by both methods: ( P F ,1.25%,24) 0.7422 using Table 5 in the rear
of the text; and ( P F ,16.076%,2) 0.7422 using the P F factor formula.
EXAMPLE 4.7
Over the past 10 years, Gentrack has placed varying sums of money into a special capital
accumulation fund. The company sells compost produced by garbage-to-compost plants in
the United States and Vietnam. Figure 4–5 is the cash flow diagram in $1000 units. Find the
amount in the account now (after 10 years) at an interest rate of 12% per year, compounded
semiannually.
Solution
Only P and F values are involved. Both methods are illustrated to find F in year 10.
Method 1: Use the semiannual CP to express the effective semiannual rate of 6% per 6-month
period. There are n (2)(number of years) semiannual periods for each cash flow. Using tabulated
factor values, the future worth by Equation [4.9] is
F 1000(FP,6%,20) 3000(FP,6%,12) 1500(FP,6%,8)
1000(3.2071) 3000(2.0122) 1500(1.5938)
$11,634 ($11.634 million)
Method 2: Express the effective annual rate, based on semiannual compounding.
Effective i% per year ( 1 0.12 ——
2 ) 2 1 12.36%
The n value is the actual number of years. Use the factor formula (FP,i,n) (1.1236) n and
Equation [4.9] to obtain the same answer as above.
F 1000(FP,12.36%,10) 3000(FP,12.36%,6) 1500(FP,12.36%,4)
1000(3.2071) 3000(2.0122) 1500(1.5938)
$11,634 ($11.634 million)
F =?
0 1 2 3 4 5 6 7 8 9
10 Year
$1000
Figure 4–5
Cash flow diagram, Example 4.7.
$3000
$1500

4.6 Equivalence Relations: Series with PP CP 109
TABLE 4–6 Examples of n and i Values Where PP CP or PP > CP
Cash Flow Series
$500 semiannually
for 5 years
$75 monthly for
3 years
$180 quarterly for
15 years
$25 per month
increase for
4 years
$5000 per quarter
for 6 years
Interest Rate
What to Find;
What Is Given Standard Notation
16% per year, Find P ; given A P 500( P A ,8%,10)
compounded
semiannually
24% per year, Find F ; given A F 75( F A ,2%,36)
compounded
monthly
5% per quarter Find F ; given A F 180( F A ,5%,60)
1% per month Find P ; given G P 25( P G ,1%,48)
1% per month Find A ; given P A 5000( A P ,3.03%,24)
4.6 Equivalence Relations: Series with PP CP
When uniform or gradient series are included in the cash flow sequence, the procedure is basically
the same as method 2 above, except that PP is now defined by the length of time between
cash flows. This also establishes the time unit of the effective interest rate. For example, if cash
flows occur on a quarterly basis, PP is 1 quarter and the effective quarterly rate is necessary. The
n value is the total number of quarters. If PP is a quarter, 5 years translates to an n value of
20 quarters. This is a direct application of the following general guideline:
When cash flows involve a series (i.e., A , G , g ) and the payment period equals or exceeds the
compounding period in length:
• Find the effective i per payment period.
• Determine n as the total number of payment periods.
In performing equivalence computations for series, only these values of i and n can be used in
interest tables, factor formulas, and spreadsheet functions. In other words, there are no other
combinations that give the correct answers, as there are for single-amount cash fl ows.
Table 4–6 shows the correct formulation for several cash flow series and interest rates. Note
that n is always equal to the total number of payment periods and i is an effective rate expressed
over the same time period as n .
EXAMPLE 4.8
For the past 7 years, Excelon Energy has paid $500 every 6 months for a software maintenance
contract. What is the equivalent total amount after the last payment, if these funds are taken
from a pool that has been returning 8% per year, compounded quarterly?
Solution
The cash flow diagram is shown in Figure 4–6. The payment period (6 months) is longer than
the compounding period (quarter); that is, PP > CP. Applying the guideline, we need to determine
an effective semiannual interest rate. Use Equation [4.7] with r 4% per 6-month
period and m 2 quarters per semiannual period.
Effective i% per 6 months ( 1 0.04 ——
2 ) 2 1 4.04%
The effective semiannual interest rate can also be obtained from Table 4–3 by using the r value
of 4% and m 2 to get i 4.04%.

110 Chapter 4 Nominal and Effective Interest Rates
The value i 4.04% seems reasonable, since we expect the effective rate to be slightly
higher than the nominal rate of 4% per 6-month period. The total number of semiannual payment
periods is n 2(7) 14. The relation for F is
F A(FA,4.04%,14)
500(18.3422)
$9171.09
To determine the FA factor value 18.3422 using a spreadsheet, enter the FV function from
Figure 2–9, that is, FV(4.04%,14,1). Alternatively, the final answer of $9171.09 can be
displayed directly using the function FV(4.04%,14,500).
i = 8% per year, compounded quarterly
F =?
0 1 2 3 4 5 6
7
Years
A = $500
Figure 4–6
Diagram of semiannual deposits used to determine F, Example 4.8.
EXAMPLE 4.9 Credit Card Offer Case
In our continuing credit card saga of Dave and his job transfer to Africa, let’s assume he did
remember that the total balance is $1030, including the $30 balance transfer fee, and he wants
to set up a monthly automatic checking account transfer to pay off the entire amount in 2 years.
Once he learned that the minimum payment is $25 per month, Dave decided to be sure the
monthly transfer exceeds this amount to avoid any further penalty fees and the penalty APR of
29.99% per year. What amount should he ask to be transferred by the due date each month?
What is the APY he will pay, if this plan is followed exactly and Chase Bank does not change
the APR during the 2-year period? Also, assume he left the credit card at home and will charge
no more to it.
Solution
The monthly A series is needed for a total of n 2(12) 24 payments. In this case,
PP CP 1 month, and the effective monthly rate is i 14.24%12 1.187% per month.
Solution by hand:
Use a calculator or hand computation to determine the A P factor value.
A P ( A P , i , n ) 1030( A P ,1.187%,24) 1030(0.04813)
$49.57 per month for 24 months
Solution by Spreadsheet: Use the function PMT(1.187%,24,1) to determine the factor
value 0.04813 to determine A for n 24 payments. Alternatively use the function
PMT (1.187%,24,1030) to directly display the required monthly payment of A $49.57.
The effective annual interest rate or APY is computed using Equation [4.7] with r 14.24%
per year, compounded monthly, and m 12 times per year.
Effective i per year ( 1 0.1424 ———
12 ) 12 – 1 1.15207 1
15.207% per year
This is the same effective annual rate i a determined in Example 4.4 b .
PE

4.6 Equivalence Relations: Series with PP CP 111
EXAMPLE 4.10
The Scott and White Health Plan (SWHP) has purchased a robotized prescription fulfillment
system for faster and more accurate delivery to patients with stable, pill-form medication for
chronic health problems, such as diabetes, thyroid, and high blood pressure. Assume this highvolume
system costs $3 million to install and an estimated $200,000 per year for all materials,
operating, personnel, and maintenance costs. The expected life is 10 years. An SWHP biomedical
engineer wants to estimate the total revenue requirement for each 6-month period that
is necessary to recover the investment, interest, and annual costs. Find this semiannual A value
both by hand and by spreadsheet, if capital funds are evaluated at 8% per year, using two different
compounding periods:
Rate 1. 8% per year, compounded semiannually .
Rate 2. 8% per year, compounded monthly .
Solution
Figure 4–7 shows the cash flow diagram. Throughout the 20 semiannual periods, the annual
cost occurs every other period, and the capital recovery series is sought for every 6-month
period. This pattern makes the solution by hand quite involved if the P F factor, not the P A
factor, is used to find P for the 10 annual $200,000 costs. The spreadsheet solution is recommended
in cases such as this.
Solution by hand—rate 1: Steps to find the semiannual A value are summarized below:
PP CP at 6 months; find the effective rate per semiannual period.
Effective semiannual i 8%2 4% per 6 months, compounded semiannually.
Number of semiannual periods n 2(10) 20.
Calculate P , using the P F factor for n 2, 4, . . . , 20 periods because the costs are annual,
not semiannual. Then use the A P factor over 20 periods to find the semiannual A .
20
P 3,000,000 200,000 [ ( P F ,4%, k ) ]
k2,4
3,000,000 200,000(6.6620) $4,332,400
A $4,332,400( A P ,4%,20) $318,778
Conclusion: Revenue of $318,778 is necessary every 6 months to cover all costs and interest
at 8% per year, compounded semiannually.
A per 6 months = ?
2 4 6 8 10 12 14 16 18
20
0
1 2 3 4 5 6 7 8 9
10
6 months
Years
$200,000 per year
i 1 = 8%, compounded semiannually
i 2 = 8%, compounded monthly
P = $3 million
Figure 4–7
Cash flow diagram with two different compounding periods, Example 4.10.

112 Chapter 4 Nominal and Effective Interest Rates
Solution by hand—rate 2: The PP is 6 months, but the CP is now monthly; therefore, PP CP.
To find the effective semiannual rate, the effective interest rate Equation [4.7] is applied with
r 4% and m 6 months per semiannual period.
Effective semiannual i ( 1 0.04 ———
6 ) 6 1 4.067%
20
P 3,000,000 200,000 [
k2,4
( P F ,4.067%, k ) ]
3,000,000 200,000(6.6204) $4,324,080
A $4,324,080( A P ,4.067%,20) $320,064
Now, $320,064, or $1286 more semiannually, is required to cover the more frequent compounding
of the 8% per year interest. Note that all P F and A P factors must be calculated
with factor formulas at 4.067%. This method is usually more calculation-intensive and errorprone
than the spreadsheet solution.
Solution by spreadsheet—rates 1 and 2: Figure 4–8 presents a general solution for the problem
at both rates. (Several rows at the bottom of the spreadsheet are not printed. They continue the
cash flow pattern of $200,000 every other 6 months through cell B32.) The functions in C8 and
E8 are general expressions for the effective rate per PP, expressed in months. This allows some
sensitivity analysis to be performed for different PP and CP values. Note the functions in C7
and E7 to determine m for the effective rate relations. This technique works well for spreadsheets
once PP and CP are entered in the time unit of the CP.
Each 6-month period is included in the cash flows, including the $0 entries, so the NPV and
PMT functions work correctly. The final A values in D14 ($318,784) and F14 ($320,069) are
the same (except for rounding) as those above.
C5/C6
E5/E6
+((1+((E2/(12/E5))/E7)) ∧ E7)1
PMT(E8,E4,F9
NPV(E8,B13:B32)+B12
Figure 4–8
Spreadsheet solution for semiannual A series for different compounding periods, Example 4.10.
4.7 Equivalence Relations: Single Amounts
and Series with PP CP
If a person deposits money each month into a savings account where interest is compounded
quarterly, do all the monthly deposits earn interest before the next quarterly compounding time?
If a person's credit card payment is due with interest on the 15th of the month, and if the full payment
is made on the 1st, does the financial institution reduce the interest owed, based on early
payment? The usual answers are no. However, if a monthly payment on a $10 million, quarterly

4.7 Equivalence Relations: Single Amounts and Series with PP CP 113
compounded, bank loan were made early by a large corporation, the corporate financial officer
would likely insist that the bank reduce the amount of interest due, based on early payment.
These are examples of PP < CP. The timing of cash flow transactions between compounding
points introduces the question of how interperiod compounding is handled. Fundamentally,
there are two policies: interperiod cash flows earn no interest, or they earn compound interest .
For a no-interperiod-interest policy, negative cash flows (deposits or payments, depending on the
perspective used for cash flows) are all regarded as made at the end of the compounding period,
and positive cash flows (receipts or withdrawals) are all regarded as made at the beginning.
As an illustration, when interest is compounded quarterly, all monthly deposits are moved to the end
of the quarter (no interperiod interest is earned), and all withdrawals are moved to the beginning (no
interest is paid for the entire quarter). This procedure can significantly alter the distribution of cash
flows before the effective quarterly rate is applied to find P , F , or A . This effectively forces the cash
flows into a PP CP situation, as discussed in Sections 4.5 and 4.6. Example 4.11 illustrates this
procedure and the economic fact that, within a one-compounding-period time frame, there is no
interest advantage to making payments early. Of course, noneconomic factors may be present.
EXAMPLE 4.11
Last year AllStar Venture Capital agreed to invest funds in Clean Air Now (CAN), a start-up
company in Las Vegas that is an outgrowth of research conducted in mechanical engineering at
the University of Nevada–Las Vegas. The product is a new filtration system used in the process
of carbon capture and sequestration (CCS) for coal-fired power plants. The venture fund manager
generated the cash flow diagram in Figure 4–9 a in $1000 units from AllStar’s perspective. Included
are payments (outflows) to CAN made over the first year and receipts (inflows) from CAN
to AllStar. The receipts were unexpected this first year; however, the product has great promise,
and advance orders have come from eastern U.S. plants anxious to become zero-emission
coal-fueled plants. The interest rate is 12% per year, compounded quarterly, and AllStar uses the
no-interperiod-interest policy. How much is AllStar in the “red” at the end of the year?
0
$150
$200
Receipts from CAN
$90
$120
0 1 2 3 4 5 6 7 8 9 10 11 12
$50
$75 $100
$90
Payments to CAN
(a)
$45
1
Year
Month
$180
$165
0
0 1 2 3
1 2 3 4
4 5 6
7 8 9 10 11 12
$50
Quarter
Month
$150
$200
$175
(b)
F = ?
Figure 4–9
( a ) Actual and ( b ) moved cash flows (in $1000) for quarterly compounding periods using no interperiod
interest, Example 4.11.

114 Chapter 4 Nominal and Effective Interest Rates
Solution
With no interperiod interest considered, Figure 4–9 b reflects the moved cash flows. All negative
cash flows (payments to CAN) are moved to the end of the respective quarter, and all
positive cash flows (receipts) are moved to the beginning of the respective quarter. Calculate
the F value at 12%4 3% per quarter.
F 1000[150( F P ,3%,4) 200( F P ,3%,3) (175 180)( F P ,3%,2)
165( F P ,3%,1) 50]
$262,111
AllStar has a net investment of $262,111 in CAN at the end of the year.
If PP CP and interperiod compounding is earned, then the cash flows are not moved, and the
equivalent P , F , or A values are determined using the effective interest rate per payment period.
The engineering economy relations are determined in the same way as in the previous two sections
for PP CP. The effective interest rate formula will have an m value less than 1, because there is
only a fractional part of the CP within one PP. For example, weekly cash flows and quarterly compounding
require that m 113 of a quarter. When the nominal rate is 12% per year, compounded
quarterly (the same as 3% per quarter, compounded quarterly), the effective rate per PP is
Effective weekly i % (1.03 ) 113 1 0.228% per week
4.8 Effective Interest Rate for Continuous
Compounding
If we allow compounding to occur more and more frequently, the compounding period becomes
shorter and shorter and m , the number of compounding periods per payment period, increases.
Continuous compounding is present when the duration of CP, the compounding period, becomes
infinitely small and m , the number of times interest is compounded per period, becomes infinite.
Businesses with large numbers of cash flows each day consider the interest to be continuously
compounded for all transactions.
As m approaches infinity, the effective interest rate Equation [4.7] must be written in a new
form. First, recall the definition of the natural logarithm base.
lim
h ( 1 1 —
h ) h e 2.71828 [4.10]
The limit of Equation [4.7] as m approaches infinity is found by using r m 1 h , which makes
m hr .
lim i lim ( 1 — r
m m m ) m 1
lim
h ( 1 1 —
h ) hr 1 lim
h [ ( 1 1 —
h ) h ] r 1
i e r 1 [4.11]
Equation [4.11] is used to compute the effective continuous interest rate, when the time periods
on i and r are the same. As an illustration, if the nominal annual r 15% per year, the effective
continuous rate per year is
i % e 0.15 1 16.183%
For convenience, Table 4–3 includes effective continuous rates for the nominal rates listed.
To find an effective or nominal interest rate for continuous compounding using the spreadsheet
functions EFFECT or NOMINAL, enter a very large value for the compounding frequency
m in Equation [4.5] or [4.6], respectively. A value of 10,000 or higher provides sufficient
accuracy. Both functions are illustrated in Example 4.12.

4.8 Effective Interest Rate for Continuous Compounding 115
EXAMPLE 4.12
(a) For an interest rate of 18% per year, compounded continuously, calculate the effective
monthly and annual interest rates.
(b) An investor requires an effective return of at least 15%. What is the minimum annual
nominal rate that is acceptable for continuous compounding?
Solution by Hand
(a) The nominal monthly rate is r 18%12 1.5%, or 0.015 per month. By Equation
[4.11], the effective monthly rate is
i % per month e r − 1 e 0.015 − 1 1.511%
Similarly, the effective annual rate using r 0.18 per year is
i % per year e r − 1 e 0.18 − 1 19.722%
(b) Solve Equation [4.11] for r by taking the natural logarithm.
e r − 1 0.15
e r 1.15
ln e r ln 1.15
r 0.13976
Therefore, a rate of 13.976% per year, compounded continuously, will generate an effective
15% per year return. The general formula to find the nominal rate, given the effective
continuous rate i , is r ln(1 i ) .
Solution by Spreadsheet
(a) Use the EFFECT function with the nominal monthly rate r 1.5% and annual rate
r 18% with a large m to display effective i values. The functions to enter on a spreadsheet
and the responses are as follows:
Monthly: EFFECT(1.5%,10000) effective i 1.511% per month
Annual: EFFECT(18%,10000) effective i 19.722% per year
(b) Use the function in Equation [4.6] in the format NOMINAL(15%,10000) to display
the nominal rate of 13.976% per year, compounded continuously.
EXAMPLE 4.13
Engineers Marci and Suzanne both invest $5000 for 10 years at 10% per year. Compute the
future worth for both individuals if Marci receives annual compounding and Suzanne receives
continuous compounding.
Solution
Marci: For annual compounding the future worth is
F P ( F P ,10%,10) 5000(2.5937) $12,969
Suzanne: Using Equation [4.11], first find the effective i per year for use in the F P factor.
Effective i % e 0.10 1 10.517%
F P ( F P ,10.517%,10) 5000(2.7183) $13,591
Continuous compounding causes a $622 increase in earnings. For comparison, daily compounding
yields an effective rate of 10.516% ( F $13,590), only slightly less than the
10.517% for continuous compounding.

116 Chapter 4 Nominal and Effective Interest Rates
For some business activities, cash flows occur throughout the day. Examples of costs are energy
and water costs, inventory costs, and labor costs. A realistic model for these activities is to
increase the frequency of the cash flows to become continuous. In these cases, the economic
analysis can be performed for continuous cash flow (also called continuous funds flow) and the
continuous compounding of interest as discussed above. Different expressions must be derived
for the factors for these cases. In fact, the monetary differences for continuous cash flows relative
to the discrete cash flow and discrete compounding assumptions are usually not large. Accordingly,
most engineering economy studies do not require the analyst to utilize these mathematical
forms to make a sound economic decision.
4.9 Interest Rates That Vary over Time
Real-world interest rates for a corporation vary from year to year, depending upon the financial
health of the corporation, its market sector, the national and international economies, forces of
inflation, and many other elements. Loan rates may increase from one year to another. Home
mortgages financed using ARM (adjustable-rate mortgage) interest is a good example. The mortgage
rate is slightly adjusted annually to reflect the age of the loan, the current cost of mortgage
money, etc.
When P , F , and A values are calculated using a constant or average interest rate over the life
of a project, rises and falls in i are neglected. If the variation in i is large, the equivalent values
will vary considerably from those calculated using the constant rate. Although an engineering
economy study can accommodate varying i values mathematically, it is more involved computationally
to do so.
To determine the P value for future cash flow values ( F t ) at different i values ( i t ) for each year t ,
we will assume annual compounding. Define
i t effective annual interest rate for year t ( t years 1 to n )
To determine the present worth, calculate the P of each F t value, using the applicable i t , and sum
the results. Using standard notation and the P F factor,
P F 1 ( P F , i 1 ,1) F 2 ( P F , i 1 ,1)( P F , i 2 ,1) . . .
F n ( P F , i 1 ,1)(P F , i 2 ,1) . . . ( P F , i n ,1) [4.12]
When only single amounts are involved, that is, one P and one F in the final year n , the last term
in Equation [4.12] is the expression for the present worth of the future cash flow.
P F n ( P F , i 1 ,1)( P F , i 2 ,1) . . . ( P F , i n ,1) [4.13]
If the equivalent uniform series A over all n years is needed, first find P , using either of the last
two equations; then substitute the symbol A for each F t symbol. Since the equivalent P has been
determined numerically using the varying rates, this new equation will have only one unknown,
namely, A . Example 4.14 illustrates this procedure.
EXAMPLE 4.14
CE, Inc., leases large earth tunneling equipment. The net profit from the equipment for each of
the last 4 years has been decreasing, as shown below. Also shown are the annual rates of return
on invested capital. The return has been increasing. Determine the present worth P and equivalent
uniform series A of the net profit series. Take the annual variation of rates of return into
account.
Year 1 2 3 4
Net Profit $70,000 $70,000 $35,000 $25,000
Annual Rate 7% 7% 9% 10%

Solution
Figure 4–10 shows the cash flows, rates for each year, and the equivalent P and A . Equation
[4.12] is used to calculate P . Since for both years 1 and 2 the net profit is $70,000 and the
annual rate is 7%, the P A factor can be used for these 2 years only.
$70,000
P [70( P A ,7%,2) 35( P F ,7%,2)( P F ,9%,1)
25( P F ,7%,2)( P F ,9%,1)( P F ,10%,1)](1000)
[70(1.8080) 35(0.8013) 25(0.7284)](1000)
$172,816 [4.14]
Chapter Summary 117
$35,000
$25,000
A = ?
0
1 2 3 4
0
1 2 3 4
Year
i = 7%
i = 7%
i = 9%
i = 10%
7% 7% 9% 10%
P = ?
$172,816
Figure 4–10
Equivalent P and A values for varying interest rates, Example 4.14.
To determine an equivalent annual series, substitute the symbol A for all net profit values
on the right side of Equation [4.14], set it equal to P $172,816, and solve for A . This
equation accounts for the varying i values each year. See Figure 4–10 for the cash flow diagram
transformation.
$172,816 A [(1.8080) (0.8013) (0.7284)] A [3.3377]
A $51,777 per year
Comment
If the average of the four annual rates, that is, 8.25% is used, the result is A $52,467. This is
a $690 per year overestimate of the equivalent annual net profit.
When there is a cash flow in year 0 and interest rates vary annually, this cash flow must be
included to determine P . In the computation for the equivalent uniform series A over all years,
including year 0, it is important to include this initial cash flow at t 0. This is accomplished by
inserting the factor value for ( P F , i 0 ,0) into the relation for A . This factor value is always 1.00. It
is equally correct to find the A value using a future worth relation for F in year n . In this case, the
A value is determined using the F P factor, and the cash flow in year n is accounted for by including
the factor ( F P , i n ,0) 1.00.
CHAPTER SUMMARY
Since many real-world situations involve cash flow frequencies and compounding periods other
than 1 year, it is necessary to use nominal and effective interest rates. When a nominal rate r is
stated, the effective interest rate i per payment period is determined by using the effective interest
rate equation.
Effective i ( 1 r —
m ) m 1

118 Chapter 4 Nominal and Effective Interest Rates
The m is the number of compounding periods (CP) per interest period. If interest compounding
becomes more and more frequent, then the length of a CP approaches zero, continuous compounding
results, and the effective i is e r 1.
All engineering economy factors require the use of an effective interest rate. The i and n values
placed in a factor depend upon the type of cash flow series. If only single amounts ( P and F )
are present, there are several ways to perform equivalence calculations using the factors. However,
when series cash flows ( A, G, and g ) are present, only one combination of the effective rate
i and number of periods n is correct for the factors. This requires that the relative lengths of PP
and CP be considered as i and n are determined. The interest rate and payment periods must have
the same time unit for the factors to correctly account for the time value of money.
From one year (or interest period) to the next, interest rates will vary. To accurately perform
equivalence calculations for P and A when rates vary significantly, the applicable interest rate
should be used, not an average or constant rate.
PROBLEMS
Nominal and Effective Rates
4.1 From the interest statement 18% per year, compounded
monthly, determine the values for interest
period, compounding period, and compounding
frequency.
4.2 From the interest statement 1% per month, determine
the values for interest period, compounding
period, and compounding frequency.
4.3 Determine the number of times interest would be
compounded in 6 months from the interest statements
( a ) 18% per year, compounded monthly,
( b ) 1% per month, and ( c ) 2% per quarter.
4.4 For an interest rate of 1% per 2 months, determine
the number of times interest would be compounded
in ( a ) 2 months, ( b ) two semiannual periods, and
( c ) 3 years.
4.5 Identify the compounding period for the following
interest statements: ( a ) 3% per quarter; ( b ) 10%
per year, compounded semiannually; ( c ) nominal
7.2% per year, compounded monthly; ( d ) effective
3.4% per quarter, compounded weekly; and ( e ) 2%
per month, compounded continuously.
4.6 Identify the following interest rate statements as
either nominal or effective: ( a ) 1.5% per month,
compounded daily; ( b ) 17% per year, compounded
quarterly; ( c ) effective 15% per year, compounded
monthly; ( d ) nominal 0.6% per month, compounded
weekly; ( e ) 0.3% per week, compounded
weekly; and ( f ) 8% per year.
4.7 Convert the given interest rates in the left-hand
column into the nominal rates listed in the righthand
column. (Assume 4 weeksmonth.)
Given Interest Rate Desired Interest Rate
1% per month Nominal rate per year
3% per quarter Nominal rate per 6 months
2% per quarter Nominal rate per year
0.28% per week Nominal rate per quarter
6.1% per 6 months Nominal rate per 2 years
4.8 What nominal interest rate per year is equivalent to
11.5% per year, compounded monthly?
4.9 For a Federal Credit Union that offers an interest
rate of 8% per year, compounded quarterly, determine
the nominal rate per 6 months.
4.10 The Second National Bank of Fullertum advertises
an APR of 14% compounded monthly for student
loans. Determine the APY. Show hand and spreadsheet
solutions.
4.11 For an effective annual rate i a of 15.87% compounded
quarterly, determine ( a ) the effective
quarterly rate and ( b ) the nominal annual rate.
( c ) What is the spreadsheet function to find the
nominal annual rate above?
4.12 High-tech companies such as IBM, AMD, and
Intel have been using nanotechnology for several
years to make microchips with faster speeds while
using less power. A less well-known company in
the chip business has been growing fast enough
that the company uses a minimum attractive rate of
return of 60% per year. If this MARR is an effective
annual rate compounded monthly, determine
the effective monthly rate.
4.13 An interest rate of 21% per year, compounded
every 4 months, is equivalent to what effective rate
per year? Show hand and spreadsheet solutions.

Problems 119
4.14 An interest rate of 8% per 6 months, compounded
monthly, is equivalent to what effective rate per
quarter?
4.15 A small company that makes modular bevel gear
drives with a tight swing ratio for optimizing pallet
truck design was told that the interest rate on a
mortgage loan would be an effective 4% per quarter,
compounded monthly. The owner was confused
by the terminology and asked you to help.
What are ( a ) the APR and ( b ) the APY?
4.16 In ‘N Out Payday Loans advertises that for a fee of
only $10, you can immediately borrow up to $200
for one month. If a person accepts the offer, what
are ( a ) the nominal interest rate per year and
( b ) the effective rate per year?
4.17 A government-required truth-in-lending document
showed that the APR was 21% and the APY was
22.71%. Determine the compounding frequency at
which the two rates are equivalent.
4.18 Julie has a low credit rating, plus she was furloughed
from her job 2 months ago. She has a new
job starting next week and expects a salary to start
again in a couple of weeks. Since she is a little
short on money to pay her rent, she decided to borrow
$100 from a loan company, which will charge
her only $10 interest if the $110 is paid no more
than 1 week after the loan is made. What are the
( a ) nominal annual and ( b ) effective annual interest
rates that she will pay on this loan?
Equivalence When PP CP
4.19 Assume you deposit 25% of your monthly check
of $5500 into a savings account at a credit union
that compounds interest semiannually. ( a ) What
are the payment and compounding periods? ( b ) Is
the payment period greater than or less than the
compounding period?
4.20 Interest is compounded quarterly, and singlepayment
cash flows (that is, F and P ) are separated
by 5 years. What must the compounding period be
on the interest rate, if the value of n in the P F or
F P equation is (a) n 5, ( b ) n 20, or ( c ) n 60?
4.21 When interest is compounded quarterly and a
uniform series cash flow of $4000 occurs every
6 months, what time periods on i and n must be
used?
4.22 Pinpoint Laser Systems is planning to set aside
$260,000 now for possibly replacing its 4-pole,
38-MW synchronous motors when it becomes
necessary. If the replacement is expected to take
place in 3 years, how much will the company have
in its investment set-aside account? Assume the
company can achieve a rate of return of 12% per
year, compounded quarterly.
4.23 Wheeling-Pittsburgh Steel is investigating whether
it should replace some of its basic oxygen furnace
equipment now or wait to do it later. The cost
later (i.e., 3 years from now) is estimated to be
$1,700,000. How much can the company afford to
spend now, if its minimum attractive rate of return
is 1.5% per month?
4.24 How much can Wells Fargo lend to a developer
who will repay the loan by selling 6 view lots at
$190,000 each 2 years from now? Assume the
bank will lend at a nominal 14% per year, compounded
semiannually.
4.25 How much will be in a high-yield account at the
National Bank of Arizona 12 years from now if
you deposit $5000 now and $7000 five years from
now? The account earns interest at a rate of 8% per
year, compounded quarterly.
4.26 Loadstar Sensors is a company that makes load
force sensors based on capacitive sensing technology.
The company wants to have $28 million for a
plant expansion 4 years from now. If the company
has already set aside $12 million in an investment
account for the expansion, how much more must
the company add to the account next year (i.e.,
1 year from now) so that it will have the $28 million
4 years from now? The account earns interest
at 12% per year, compounded quarterly.
4.27 A structural engineering consulting company is
examining its cash flow requirements for the next
6 years. The company expects to replace office
machines and computer equipment at various
times over the 6-year planning period. Specifically,
the company expects to spend $21,000 two
years from now, $24,000 three years from now,
and $10,000 five years from now. What is the present
worth of the planned expenditures at an interest
rate of 10% per year, compounded semiannually?
4.28 Irvin Aerospace of Santa Ana, California, was
awarded a 5-year contract to develop an advanced
space capsule airbag landing attenuation system
for NASA’s Langley Research Center. The company’s
computer system uses fluid structure interaction
modeling to test and analyze each airbag
design concept. What is the present worth of the
contract at 16% per year, compounded quarterly, if
the quarterly cost in years 1 through 5 is $2 million
per quarter?

120 Chapter 4 Nominal and Effective Interest Rates
4.29 Heyden Motion Solutions ordered $7 million
worth of seamless tubes for its drill collars from
the Timken Company of Canton, Ohio. (A drill
collar is the heavy tubular connection between a
drill pipe and a drill bit.) At 12% per year, compounded
semiannually, what is the equivalent uniform
cost per semiannual period over a 5-year
amortization period?
4.30 In 2010, the National Highway Traffic Safety Administration
raised the average fuel efficiency standard
to 35.5 miles per gallon (mpg) for cars and
light trucks by the year 2016. The rules will cost
consumers an average of $926 extra per vehicle in
the 2016 model year. Assume Yolanda will purchase
a new car in 2016 and plans to keep it for
5 years. How much will the monthly savings in the
cost of gasoline have to be to recover Yolanda’s
extra cost? Use an interest rate of 0.75% per month.
4.31 Fort Bliss, a U.S. Army military base, contributed
$3.3 million of the $87 million capital cost for a
desalting plant constructed and operated by El Paso
Water Utilities (EPWU). In return, EPWU agreed
to sell water to Fort Bliss at $0.85 per 1000 gallons
for 20 years. If the army base uses 200 million gallons
of water per month, what is the Army’s cost
per month for water? The $3.3 million capital cost
is amortized at an interest rate of 6% per year,
compounded monthly.
4.32 Beginning in 2011, city hall, administrative offices,
and municipal courts in the city of El Paso,
Texas, will go on a 10 hourday, 4-day workweek
from the beginning of May through the end of September.
The shortened workweek will affect 25%
of the city’s 6100 employees and will save $42,600
per month for those 5 months, because of reduced
fuel, utility, and janitorial expenses. If this work
schedule continues for the next 10 years, what is
the future worth of the savings at the end of that
time (i.e., end of year 2020)? Use an interest rate
of 0.5% per month.
4.33 In October 2009, Wal-Mart started selling caskets
on its website that undercut many funeral homes.
Prices ranged from $999 for steel models such as
Dad Remembered to $3199 for the Sienna Bronze
casket. Part of the business model is to get people
to plan ahead, so the company is allowing people
to pay for the caskets over a 12-month period with
no interest. An individual purchased a Sienna
Bronze casket and made 12 equal monthly payments
(in months 1 through 12) at no interest. How
much did this person save each month compared to
another person who paid an interest rate of 6% per
year, compounded monthly?
4.34 NRG Energy plans to construct a giant solar plant
in Santa Teresa, New Mexico, to supply electricity
to West Texas Electric. The plant will have 390,000
heliostats to concentrate sunlight onto 32 water
towers to generate steam. It will provide enough
power for 30,000 homes. If the company spends
$28 million per month for 2 years in constructing
the plant, how much will the company have to
make each month in years 3 through 22 (that is,
20 years) to recover its investment plus 18% per
year, compounded monthly?
4.35 Many college students have Visa credit cards that
carry an interest rate of “simple 24% per year”
(that is, 2% per month). When the balance on
such a card is $5000, the minimum payment is
$110.25.
(a) What is the amount of interest in the first
payment?
(b) How long will it take, in months, to pay off
the balance, if the cardholder continues to
make payments of $110.25 per month and
adds no other charges to the card?
4.36 Bart is an engineering graduate who did not take
the engineering economy elective during his B.S.
degree course work. After working for a year or so,
he found himself in financial trouble, and he borrowed
$500 from a friend in the finance department
at his office. Bart agreed to repay the loan
principal plus $75 interest 1 month later. The two
got separated doing different jobs, and 1 year went
by. The friend e-mailed Bart after exactly 1 year
and asked for the loan repayment plus the interest
with monthly compounding, since Bart had made
no effort to repay the loan during the year.
( a ) What does Bart now owe his friend? ( b ) What
effective annual interest rate did Bart pay on this
$500 loan?
4.37 AT&T announced that the early termination fee for
smart phones will jump from $175 to $375. The
fee will be reduced by $10 each month of the
2-year contract. Wireless carriers justify the fees
by pointing out that the cost of a new phone is
heavily discounted from what the carrier pays the
manufacturer. Assume AT&T pays $499 for an
iPhone that it sells for $199. How much profit
would the company have to make each month (i.e.,
prior to the termination), if it wanted to make a rate
of return of 1.5% per month on its $300 investment
in a customer who terminates the contract
after 12 months?
4.38 What is the future worth of a present cost of
$285,000 to Monsanto, Inc. 5 years from now at an
interest rate of 2% per month?

Problems 121
4.39 Environmental recovery company RexChem Partners
plans to finance a site reclamation project
that will require a 4-year cleanup period. The
company plans to borrow $3.6 million now. How
much will the company have to get in a lump-sum
payment when the project is over in order to earn
24% per year, compounded quarterly, on its investment?
4.40 For the cash flows shown below, determine the
equivalent uniform worth in years 1 through 5 at an
interest rate of 18% per year, compounded monthly.
Year 1 2 3 4 5
Cash Flow, $ 200,000 0 350,000 0 400,000
4.41 For the cash flow diagram shown, solve for F ,
using an interest rate of 1% per month.
0
1
i = 1% per month
2 3 4 5 6 7 8
$30 $30 $30 $30 $30 $30
F =?
$50 $50 $50
Years
4.42 According to the Government Accountability Office
(GAO), if the U.S. Postal Service does not
change its business model, it will lose $480 million
next month and $500 million the month after that,
and the losses will increase by $20 million per
month for the next 10 years. At an interest rate of
0.25% per month, what is the equivalent uniform
amount per month of the losses through year 10?
4.43 Equipment maintenance costs for manufacturing
explosion-proof pressure switches are projected to
be $100,000 in year 1 and increase by 4% each
year through year 5. What is the present worth of
the maintenance costs at an interest rate of 10%
per year, compounded quarterly?
Equivalence When PP CP
4.44 If you deposit $1000 per month into an investment
account that pays interest at a rate of 6% per year,
compounded quarterly, how much will be in the
account at the end of 5 years? There is no interperiod
compounding.
4.45 Freeport McMoran purchased two model MTVS
peristaltic pumps (to inject sulfuric acid and King
Lee antiscalant) for use at its nanofiltration water
conditioning plant. The cost of the pumps was
$950 each. If the chemical cost is $11 per day,
determine the equivalent cost per month at an interest
rate of 12% per year, compounded monthly.
Assume 30 days per month and a 3-year pump
life.
4.46 Income from recycling paper and cardboard at the
U.S. Army's Fort Benning Maneuver Center has
averaged $3000 per month for 2½ years. What is
the future worth of the income (after the 2½ years)
at an interest rate of 6% per year, compounded
quarterly? Assume there is no interperiod compounding.
4.47 The Autocar E3 refuse truck has an energy recovery
system developed by Parker Hannifin
LLC that is expected to reduce fuel consumption
by 50%. Pressurized fluid flows from carbon
fi ber-reinforced accumulator tanks to two hydrostatic
motors that propel the vehicle forward.
(The truck recharges the accumulators when it
brakes.) The fuel cost for a regular refuse truck is
$900 per month. How much can a private wastehauling
company afford to spend now on the
recovery system, if it wants to recover its investment
in 3 years plus a return of 14% per year,
compounded semiannually? Assume no interperiod
compounding.
Continuous Compounding
4.48 What effective interest rate per year is equal to
1.2% per month, compounded continuously? Show
hand and spreadsheet solutions.
4.49 What effective interest rate per quarter is equal to
a nominal 1.6% per month, compounded continuously?
4.50 What nominal rate per month is equivalent to an
effective 1.3% per month, compounded continuously?
4.51 Companies such as GE that have huge amounts of
cash flow every day base their financial calculations
on continuous compounding. If the company wants
to make an effective 25% per year, compounded
continuously, what nominal daily rate of return has
to be realized? Assume 365 days per year.
4.52 U.S. Steel is planning a plant expansion that is expected
to cost $13 million. How much money must
the company set aside now in a lump-sum investment
to have the money in 2 years? Capital funds
earn interest at a rate of 12% per year, compounded
continuously.

122 Chapter 4 Nominal and Effective Interest Rates
4.53 Periodic outlays for inventory control software at
Baron Chemicals are expected to be $150,000
immediately, $200,000 in 1 year, and $350,000 in
2 years. What is the present worth of the costs at
an interest rate of 10% per year, compounded
continuously?
Varying Interest Rates
4.54 Many small companies use accounts receivable as
collateral to borrow money for continuing operations
and meeting payrolls. If a company borrows
$300,000 now at an interest rate of 1% per month,
but the rate changes to 1.25% per month after
4 months, how much will the company owe at the
end of 1 year?
4.55 The maintenance cost for furnaces at a copper
smelting plant have been constant at $140,000 per
year for the past 5 years. If the interest rate was 8%
per year for the first 3 years and then it increased to
10% in years 4 and 5, what is the equivalent future
worth (in year 5) of the maintenance cost? Show
hand and spreadsheet solutions.
4.56 By filling carbon nanotubes with miniscule wires
made of iron and iron carbide, incredibly thin
nanowires can be extruded by blasting the carbon
nanotubes with an electron beam. If Gentech Technologies
plans to spend $1.7 million in year 1,
$2.1 million in year 2, and $3.4 million in year 3 to
develop the technology, determine the present
worth of the investments in year 0, if the interest
rate in year 1 is 10% and in years 2 and 3 it is 12%
per year.
4.57 Find ( a ) the present worth P , and ( b ) the equivalent
uniform annual worth A for the cash flows
shown below.
P =?
0
1
i = 10% i = 14%
2 3 4 5 6 7 8
$100 $100 $100 $100 $100
$160 $160 $160
Year
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
4.58 The term annual percentage rate is the same as:
( c) 6.0% per quarter, compounded continuously
(a) Effective rate
(d) 9% per 6 months
(b) Nominal rate
(c) Annual percentage yield
4.62 An effective 12.68% per year, compounded monthly,
(d) All of the above
is the closest to:
(a) 12% per year
4.59 An interest rate is an effective rate under all of the
following conditions, except when:
(b)
(c)
12% per year, compounded annually
1% per month
(a) The compounding period is not stated
(d) 1% per month, compounded annually
(b) The interest period and compounding period
are the same
(c) The interest statement says that the interest
rate is effective
(d) The interest period is shorter than the compounding
period
4.60 An interest rate of nominal 12% per year, compounded
weekly, is:
(a) An effective rate per year
(b) An effective rate per week
(c) A nominal rate per year
(d) A nominal rate per week
4.61 An interest rate of 1.5% per month, compounded
continuously, is the same as:
(a) An effective 1.5% per month
( b) 4.5% per quarter, compounded continuously
4.63 For an interest rate of 2% per month, the effective
semiannual interest rate is:
( a ) 2.02%
( b ) 12.005%
( c ) 12.31%
( d ) 12.62%
4.64 If you make quarterly deposits for 3 years into an
account that compounds interest at 1% per month,
the value of n in the F A factor that will determine
F at the end of the 3-year period is:
( a ) 3 ( b ) 12
( c ) 36 ( d ) None of these

Additional Problems and FE Exam Review Questions 123
4.65 Identify the following interest rates as nominal or
effective.
Rate 1: 1.5% per quarter
Rate 2: 1.5% per quarter, compounded
monthly
( a) Both are nominal rates.
( b) Rate 1 is nominal and rate 2 is effective.
( c) Rate 1 is effective and rate 2 is nominal.
( d) Both are effective.
4.66 In solving uniform series problems, n in the standard
factor notation equation is equal to the
number of arrows (i.e., cash flows) in the original
cash flow diagram when:
( a) The payment period (PP) is longer than the
compounding period (CP).
( b) The compounding period is equal to the payment
period.
( c) Both ( a ) and ( b ) are correct.
( d) The compounding period is longer than the
payment period.
4.67 An engineer who is saving for her retirement plans
to deposit $500 every quarter, starting one quarter
from now, into an investment account. If the account
pays interest at 6% per year, compounded
semiannually, the total she will have at the end of
25 years is closest to:
( a ) $50,000 ( b ) $56,400
( c ) $79,700 ( d ) $112,800
4.68 A company that makes flange-mount, motorized rotary
potentiometers expects to spend $50,000 for a
certain machine 4 years from now. At an interest rate
of 12% per year, compounded quarterly, the present
worth of the machine’s cost is represented by the following
equation:
( a) P 50,000( P F , 3%, 16)
( b) P 50,000( P F , effective i 6 months, 8)
( c) P 50,000( P F , effective i year, 4)
( d) Any of the above
4.69 For the cash flow diagram shown, the unit of the
payment period (PP) is:
( a ) Months ( b ) Quarters
( c ) Semiannual ( d ) Years
0
1
i = 1% per month
2 3 4 5 6 7 8 9 10 11 12
1 2
$500
$480
$460
$400 $400
$420
$440
F =?
3
Quarter
Year
4.70 A small company plans to spend $10,000 in year 2
and $10,000 in year 5. At an interest rate of effective
10% per year, compounded semiannually, the equation
that represents the equivalent annual worth A in
years 1 through 5 is:
( a) A 10,000( P F ,10%,2)(A P ,10%,5)
10,000( A F ,10%,5)
(b) A 10,000( A P ,10%,4) 10,000( A F ,10%,5)
(c) A 10,000( P F ,5%,2)( A P ,5%,10)
10,000( A F ,5%,10)
(d) A [10,000( F P ,10%,5)
10,000]( A F ,10%,5)
4.71 Assume you make monthly deposits of $200 starting
one month from now into an account that pays
6% per year, compounded semiannually. If you
want to know the total after 4 years , the value of n
you should use in the F A factor is:
( a ) 2 ( b ) 4 ( c ) 8 ( d ) 12
4.72 The cost of replacing part of a cell phone videochip
production line in 6 years is estimated to be
$500,000. At an interest rate of 14% per year,
compounded semiannually, the uniform amount
that must be deposited into a sinking fund every
6 months is closest to:
( a ) $21,335 ( b ) $24,825
( c ) $27,950 ( d ) $97,995

124 Chapter 4 Nominal and Effective Interest Rates
CASE STUDY
IS OWNING A HOME A NET GAIN OR NET LOSS OVER TIME?
Background
The Carroltons are deliberating whether to purchase a house
or continue to rent for the next 10 years. They are assured by
both of their employers that no transfers to new locations will
occur for at least this number of years. Plus, the high school
that their children attend is very good for their college prep
education, and they all like the neighborhood where they
live now.
They have a total of $40,000 available now and estimate
that they can afford up to $2850 per month for the total house
payment.
If the Carroltons do not buy a house, they will continue
to rent the house they currently occupy for $2700 per
month. They will also place the $40,000 into an investment
instrument that is expected to earn at the rate of 6% per
year. Additionally, they will add to this investment at the
end of each year the same amount as the monthly 15-year
mortgage payments. This alternative is called the rent–
don’t buy plan.
Information
Two financing plans using fixed-rate mortgages are currently
available. The details are as follows.
Plan
A
B
Description
30-year fixed rate of 5.25% per year interest;
10% down payment
15-year fixed rate of 5.0% per year interest;
10% down payment
Other information:
• Price of the house is $330,000 .
• Taxes and insurance (T&I) are $500 per month.
• Up-front fees (origination fee, survey fee, attorney’s fee,
etc.) are $3000.
Any money not spent on the down payment or monthly payment
will be invested and return at a rate of 6% per year
(0.5% per month).
The Carroltons anticipate selling the house after 10 years
and plan for a 10% increase in price, that is, $363,000 (after
all selling expenses are paid)
Case Study Exercises
1. The 30-year fixed-rate mortgage (plan A) is analyzed
below. No taxes are considered on proceeds from the
savings or investments.
Perform a similar analysis for the 15-year loan
(plan B) and the rent–don’t buy plan. The Carroltons
decided to use the largest future worth after 10 years to
select the best of the plans. Do the analysis for them and
select the best plan.
Plan A analysis: 30-year fixed-rate loan
Amount of money required for closing costs:
Down payment (10% of $330,000) $33,000
Up-front fees (origination fee, attorney’s
fee, survey, filing fee, etc.) 3,000
Total $36,000
The amount of the loan is $297,000, and equivalent
monthly principal and interest (P&I) is determined at
5.25%12 0.4375% per month for 30(12) 360 months.
A 297,000( A P ,0.4375%,360) 297,000(0.005522)
$1640
Add the T&I of $500 for a total monthly payment of
Payment A $2140 per month
The future worth of plan A is the sum of three future
worth components: remainder of the $40,000 available
for the closing costs ( F 1 A ); left-over money from that
available for monthly payments ( F 2 A ); and increase in
the house value when it is sold after 10 years ( F 3 A ).
These are calculated here.
F 1 A (40,000 36,000)( F P ,0.5%,120)
$7278
Money available each month to invest after the mortgage
payment, and the future worth after 10 years is
2850 2140 $710
F 2 A 710( F A ,0.5%,120)
$116,354

Case Study 125
Net money from the sale in 10 years ( F 3 A ) is the difference
between the net selling price ($363,000) and the
remaining balance on the loan.
Loan balance
297,000( F P ,0.4375%,120)
1640( F A ,0.4375%,120)
297,000(1.6885) 1640(157.3770)
$243,386
F 3 A 363,000 243,386 $119,614
Total future worth of plan A is
F A F 1 A F 2 A F 3 A
7278 116,354 119,614
$243,246
2. Perform this analysis if all estimates remain the same,
except that when the house sells 10 years after purchase,
the bottom has fallen out of the housing market and the
net selling price is only 70% of the purchase price, that
is, $231,000.

LEARNING STAGE 2
Basic Analysis Tools
LEARNING STAGE 2
Basic Analysis
Tools
CHAPTER 5
Present Worth
Analysis
CHAPTER 6
Annual Worth
Analysis
CHAPTER 7
Rate of Return
Analysis: One Project
CHAPTER 8
Rate of Return
Analysis: Multiple
Alternatives
CHAPTER 9
BenefitCost Analysis
and Public Sector
Economics
An engineering project or alternative is formulated to make or
purchase a product, to develop a process, or to provide a
service with specified results. An engineering economic
analysis evaluates cash flow estimates for parameters such as initial
cost, annual costs and revenues, nonrecurring costs, and possible
salvage value over an estimated useful life of the product; process,
or service. The chapters in this Learning Stage develop and demonstrate
the basic tools and techniques to evaluate one or more alternatives
using the factors, formulas, and spreadsheet functions
learned in Stage 1.
After completing these chapters, you will be able to evaluate
most engineering project proposals using a well-accepted economic
analysis technique, such as present worth, future worth, capitalized
cost, life-cycle costing, annual worth, rate of return, or benefit /cost
analysis.
The epilogue to this stage provides an approach useful in selecting
the engineering economic method that will provide the best
analysis for the estimates and conditions present once the mutually
exclusive alternatives are defined.
Important note: If depreciation andor after-tax analysis is to be
considered along with the evaluation methods in Chapters 5 through
9, Chapter 16 and/or Chapter 17 should be covered, preferably after
Chapter 6.

CHAPTER 5
Present Worth
Analysis
LEARNING OUTCOMES
Purpose: Utilize different present worth techniques to evaluate and select alternatives.
SECTION TOPIC LEARNING OUTCOME
5.1 Formulate alternatives • Identify mutually exclusive and independent
projects; define revenue and cost alternatives.
5.2 PW of equal-life alternatives • Select the best of equal-life alternatives using
present worth analysis.
5.3 PW of different-life alternatives • Select the best of different-life alternatives
using present worth analysis.
5.4 FW analysis • Select the best alternative using future worth
analysis.
5.5 CC analysis • Select the best alternative using capitalized cost
(CC) analysis.

A
future amount of money converted to its equivalent value now has a present worth
(PW) that is always less than that of the future cash flow, because all P F factors have
a value less than 1.0 for any interest rate greater than zero. For this reason, present
worth values are often referred to as discounted cash flows (DCF), and the interest rate is referred
to as the discount rate. Besides PW, two other terms frequently used are present value
( PV ) and net present value ( NPV ). Up to this point, present worth computations have been
made for one project or alternative. In this chapter, techniques for comparing two or more
mutually exclusive alternatives by the present worth method are treated. Two additional applications
are covered here—future worth and capitalized cost. Capitalized costs are used for
projects with very long expected lives or long planning horizons.
To understand how to organize an economic analysis, this chapter begins with a description
of independent and mutually exclusive projects as well as revenue and cost
alternatives.
Water for Semiconductor Manufacturing
Case: The worldwide contribution of
semiconductor sales is about $250 billion
per year, or about 10% of the world’s
GDP (gross domestic product). This industry
produces the microchips used in many
of the communication, entertainment,
transportation, and computing devices
we use every day. Depending upon the
type and size of fabrication plant (fab),
the need for ultrapure water (UPW) to
manufacture these tiny integrated circuits
is high, ranging from 500 to 2000 gpm
(gallons per minute). Ultrapure water is
obtained by special processes that commonly
include reverse osmosis deionizing
resin bed technologies. Potable water
obtained from purifying seawater or
brackish groundwater may cost from
$2 to $3 per 1000 gallons, but to obtain
UPW on-site for semiconductor manufacturing
may cost an additional $1 to $3 per
1000 gallons.
A fab costs upward of $2.5 billion to
construct, with approximately 1% of this
total, or $25 million, required to provide
the ultrapure water needed, including
the necessary wastewater and recycling
equipment.
A newcomer to the industry, Angular
Enterprises, has estimated the cost profiles
for two options to supply its anticipated
fab with water. It is fortunate to
PE
have the option of desalinated seawater
or purified groundwater sources in the
location chosen for its new fab. The initial
cost estimates for the UPW system are
given below.
Source
Equipment first
cost, $M
Seawater
(S)
20
Groundwater
(G)
22
AOC, $M per year 0.5 0.3
Salvage value, % of
first cost
Cost of UPW, $ per
1000 gallons
5 10
4 5
Angular has made some initial estimates
for the UPW system.
Life of UPW equipment 10 years
UPW needs
Operating time
1500 gpm
16 hours per
day for 250 days
per year
This case is used in the following topics
(Sections) and problems of this chapter:
PW analysis of equal-life alternatives
(Section 5.2)
PW analysis of different-life alternatives
(Section 5.3)
Capitalized cost analysis (Section 5.5)
Problems 5.20 and 5.34
5.1 Formulating Alternatives
The evaluation and selection of economic proposals require cash flow estimates over a stated
period of time, mathematical techniques to calculate the measure of worth (review Example
1.2 for possible measures), and a guideline for selecting the best proposal. From all the

130 Chapter 5 Present Worth Analysis
Mandates
Ideas
Information
Experience
Estimates
Plans
Proposals
Not viable
Viable
Not viable
A
B
C
1
…
m
2
D
E
Mutually
exclusive
alternatives
Select
only
one
1
2
…
m
+
DN
Either
of these
Nature
of proposals
Independent
projects
Select
all
justified
DN
1
2
…
m
DN = do nothing
Types of cash flow estimates
* Revenues and costs
* Costs only
Revenue alternative
Cost alternative
Perform evaluation and make selection
Figure 5–1
Progression from proposals to economic evaluation to selection.
proposals that may accomplish a stated purpose, the alternatives are formulated. This progression
is detailed in Figure 5–1. Up front, some proposals are viable from technological, economic,
andor legal perspectives; others are not viable. Once the obviously nonviable ideas
are eliminated, the remaining viable proposals are fleshed out to form the alternatives to be
evaluated. Economic evaluation is one of the primary means used to select the best
alternative(s) for implementation.
The nature of the economic proposals is always one of two types:
Mutually exclusive alternatives: Only one of the proposals can be selected. For terminology
purposes, each viable proposal is called an alternative.
Independent projects: More than one proposal can be selected. Each viable proposal is called a
project.
The do-nothing (DN) proposal is usually understood to be an option when the evaluation is
performed.

5.2 Present Worth Analysis of Equal-Life Alternatives 131
The DN alternative or project means that the current approach is maintained; nothing new
is initiated. No new costs, revenues, or savings are generated.
Do nothing
If it is absolutely required that one or more of the defined alternatives be selected, do nothing is
not considered. This may occur when a mandated function must be installed for safety, legal,
government, or other purposes.
Mutually exclusive alternatives and independent projects are selected in completely different
ways. A mutually exclusive selection takes place, for example, when an engineer must select
the best diesel-powered engine from several available models. Only one is chosen, and the
rest are rejected. If none of the alternatives are economically justified, then all can be rejected
and, by default, the DN alternative is selected. For independent projects one, two or more, in
fact, all of the projects that are economically justified can be accepted, provided capital funds
are available. This leads to the two following fundamentally different evaluation bases:
Mutually exclusive alternatives compete with one another and are compared pairwise.
Independent projects are evaluated one at a time and compete only with the DN project.
Any of the techniques in Chapters 5 through 9 can be used to evaluate either type of proposal—
mutually exclusive or independent. When performed correctly as described in each chapter, any
of the techniques will reach the same conclusion of which alternative or alternatives to select.
This chapter covers the present worth method.
A parallel can be developed between independent and mutually exclusive evaluation. Assume
there are m independent projects. Zero, one, two, or more may be selected. Since each project may
be in or out of the selected group of projects, there are a total of 2 m mutually exclusive alternatives.
This number includes the DN alternative, as shown in Figure 5–1. For example, if the engineer
has three diesel engine models (A, B, and C) and may select any number of them, there are
2 3 8 alternatives: DN, A, B, C, AB, AC, BC, ABC. Commonly, in real-world applications, there
are restrictions, such as an upper budgetary limit, that eliminate many of the 2 m alternatives. Independent
project analysis without budget limits is discussed in this chapter and through Chapter 9.
Chapter 12 treats independent projects with a budget limitation; this is called capital budgeting.
Finally, it is important to recognize the nature of the cash flow estimates before starting the
computation of a measure of worth that leads to the final selection. Cash flow estimates determine
whether the alternatives are revenue- or cost-based. All the alternatives or projects must be
of the same type when the economic study is performed. Definitions for these types follow:
Revenue: Each alternative generates cost (cash outflow) and revenue (cash inflow) estimates,
and possibly savings, also considered cash inflows. Revenues can vary for each alternative.
Cost: Each alternative has only cost cash flow estimates. Revenues or savings are assumed
equal for all alternatives; thus they are not dependent upon the alternative selected. These are
also referred to as service alternatives.
Revenue or cost
alternative
Although the exact procedures vary slightly for revenue and cost cash flows, all techniques and
guidelines covered through Chapter 9 apply to both. Differences in evaluation methodology are
detailed in each chapter.
5.2 Present Worth Analysis of Equal-Life Alternatives
The PW comparison of alternatives with equal lives is straightforward. The present worth P is
renamed PW of the alternative. The present worth method is quite popular in industry because all
future costs and revenues are transformed to equivalent monetary units NOW; that is, all future
cash flows are converted (discounted) to present amounts (e.g., dollars) at a specific rate of return,
which is the MARR. This makes it very simple to determine which alternative has the best economic
advantage. The required conditions and evaluation procedure are as follows:
If the alternatives have the same capacities for the same time period (life), the equal-service
requirement is met. Calculate the PW value at the stated MARR for each alternative.
Equal-service
requirement

132 Chapter 5 Present Worth Analysis
For mutually exclusive (ME) alternatives, whether they are revenue or cost alternatives, the following
guidelines are applied to justify a single project or to select one from several alternatives.
Project evaluation
ME alternative
selection
One alternative: If PW 0, the requested MARR is met or exceeded and the alternative is
economically justified.
Two or more alternatives: Select the alternative with the PW that is numerically largest,
that is, less negative or more positive. This indicates a lower PW of cost for cost alternatives
or a larger PW of net cash flows for revenue alternatives.
Note that the guideline to select one alternative with the lowest cost or highest revenue uses
the criterion of numerically largest. This is not the absolute value of the PW amount, because
the sign matters. The selections below correctly apply the guideline for two alternatives A and B.
PW A PW B Selected Alternative
$2300 $1500 B
500 1000 B
2500 2000 A
4800 400 A
For independent projects, each PW is considered separately, that is, compared with the DN
project, which always has PW 0. The selection guideline is as follows:
Independent project
selection
One or more independent projects: Select all projects with PW 0 at the MARR.
The independent projects must have positive and negative cash flows to obtain a PW value that
can exceed zero; that is, they must be revenue projects.
All PW analyses require a MARR for use as the i value in the PW relations. The bases used
to establish a realistic MARR were summarized in Chapter 1 and are discussed in detail in
Chapter 10.
EXAMPLE 5.1
A university lab is a research contractor to NASA for in-space fuel cell systems that are hydrogenand
methanol-based. During lab research, three equal-service machines need to be evaluated
economically. Perform the present worth analysis with the costs shown below. The MARR is
10% per year.
Electric-Powered Gas-Powered Solar-Powered
First cost, $ 4500 3500 6000
Annual operating cost (AOC), $/year 900 700 50
Salvage value S, $ 200 350 100
Life, years 8 8 8
Solution
These are cost alternatives. The salvage values are considered a “negative” cost, so a sign
precedes them. (If it costs money to dispose of an asset, the estimated disposal cost has a sign.)
The PW of each machine is calculated at i 10% for n 8 years. Use subscripts E , G , and S .
PW E 4500 900( P A ,10%,8) 200( P F ,10%,8) $9208
PW G 3500 700( P A ,10%,8) 350( P F ,10%,8) $7071
PW S 6000 50( P A ,10%,8) 100( P F ,10%,8) $6220
The solar-powered machine is selected since the PW of its costs is the lowest; it has the
numerically largest PW value.

5.3 Present Worth Analysis of Different-Life Alternatives 133
EXAMPLE 5.2 Water for Semiconductor Manufacturing Case
As discussed in the introduction to this chapter, ultrapure water (UPW) is an expensive commodity
for the semiconductor industry. With the options of seawater or groundwater sources,
it is a good idea to determine if one system is more economical than the other. Use a MARR of
12% per year and the present worth method to select one of the systems.
PE
Solution
An important first calculation is the cost of UPW per year. The general relation and estimated
costs for the two options are as follows:
$
UPW cost relation: ——
year cost in $
——————
(
———
1000 gallons ) ( gallons ————
minute ) ( minutes
hour
Seawater: (41000)(1500)(60)(16)(250) $1.44 M per year
Groundwater: (51000)(1500)(60)(16)(250) $1.80 M per year
) ( hours ———
day ) (——
days
year )
Calculate the PW at i 12% per year and select the option with the lower cost (larger PW
value). In $1 million units:
PW relation: PW first cost PW of AOC PW of UPW PW of salvage value
PW S 20 0.5(PA,12%,10) 1.44(PA,12%,10) 0.05(20)(PF,12%,10)
20 0.5(5.6502) 1.44(5.6502) 1(0.3220)
$30.64
PW G 22 0.3(PA,12%,10) 1.80(PA,12%,10) 0.10(22)(PF,12%,10)
22 0.3(5.6502) 1.80(5.6502) 2.2(0.3220)
$33.16
Based on this present worth analysis, the seawater option is cheaper by $2.52 M.
5.3 Present Worth Analysis of Different-Life
Alternatives
When the present worth method is used to compare mutually exclusive alternatives that have
different lives, the equal-service requirement must be met. The procedure of Section 5.2 is followed,
with one exception:
The PW of the alternatives must be compared over the same number of years and must end
at the same time to satisfy the equal-service requirement.
This is necessary, since the present worth comparison involves calculating the equivalent PW of
all future cash flows for each alternative. A fair comparison requires that PW values represent
cash flows associated with equal service. For cost alternatives, failure to compare equal service
will always favor the shorter-lived mutually exclusive alternative, even if it is not the more economical
choice, because fewer periods of costs are involved. The equal-service requirement is
satisfied by using either of two approaches:
LCM: Compare the PW of alternatives over a period of time equal to the least common
multiple (LCM) of their estimated lives.
Study period: Compare the PW of alternatives using a specified study period of n years.
This approach does not necessarily consider the useful life of an alternative. The study period
is also called the planning horizon.
Equal-service
requirement
LCM or study
period
For either approach, calculate the PW at the MARR and use the same selection guideline as
that for equal-life alternatives. The LCM approach makes the cash flow estimates extend to the
same period, as required. For example, lives of 3 and 4 years are compared over a 12-year period.

134 Chapter 5 Present Worth Analysis
The first cost of an alternative is reinvested at the beginning of each life cycle, and the estimated
salvage value is accounted for at the end of each life cycle when calculating the PW values over
the LCM period. Additionally, the LCM approach requires that some assumptions be made about
subsequent life cycles.
The assumptions when using the LCM approach are that
1. The service provided will be needed over the entire LCM years or more.
2. The selected alternative can be repeated over each life cycle of the LCM in exactly the same
manner.
3. Cash flow estimates are the same for each life cycle.
As will be shown in Chapter 14, the third assumption is valid only when the cash flows are
expected to change by exactly the inflation (or deflation) rate that is applicable through the LCM
time period. If the cash flows are expected to change by any other rate, then the PW analysis must
be conducted using constant-value dollars, which considers inflation (Chapter 14).
A study period analysis is necessary if the first assumption about the length of time the alternatives
are needed cannot be made. For the study period approach, a time horizon is chosen over
which the economic analysis is conducted, and only those cash flows which occur during that time
period are considered relevant to the analysis. All cash flows occurring beyond the study period are
ignored. An estimated market value at the end of the study period must be made. The time horizon
chosen might be relatively short, especially when short-term business goals are very important. The
study period approach is often used in replacement analysis (Chapter 11). It is also useful when the
LCM of alternatives yields an unrealistic evaluation period, for example, 5 and 9 years.
EXAMPLE 5.3
National Homebuilders, Inc., plans to purchase new cut-and-finish equipment. Two manufacturers
offered the estimates below.
Vendor A
Vendor B
First cost, $ 15,000 18,000
Annual M&O cost, $ per year 3,500 3,100
Salvage value, $ 1,000 2,000
Life, years 6 9
(a) Determine which vendor should be selected on the basis of a present worth comparison,
if the MARR is 15% per year.
(b) National Homebuilders has a standard practice of evaluating all options over a 5-year
period. If a study period of 5 years is used and the salvage values are not expected to
change, which vendor should be selected?
Solution
(a) Since the equipment has different lives, compare them over the LCM of 18 years. For life
cycles after the first, the first cost is repeated in year 0 of each new cycle, which is the last
year of the previous cycle. These are years 6 and 12 for vendor A and year 9 for B. The
cash flow diagram is shown in Figure 5–2. Calculate PW at 15% over 18 years.
PW A 15,000 15,000(PF,15%,6) 1000(PF,15%,6)
15,000(PF,15%,12) 1000(PF,15%,12) 1000(PF,15%,18)
3,500(PA,15%,18)
$45,036
PW B 18,000 18,000(PF,15%,9) 2000(PF,15%,9)
2000(PF,15%,18) 3100(PA,15%,18)
$41,384

5.3 Present Worth Analysis of Different-Life Alternatives 135
PW A =?
$1000 $1000
$1000
1 2 6 12 16 17 18
Year
$3500
$15,000
$15,000
Vendor A
$15,000
PW B =?
$2000
$2000
1 2 9
16 17 18
Year
$3100
$18,000
$18,000
Vendor B
Figure 5–2
Cash flow diagram for different-life alternatives, Example 5.3 a .
Vendor B is selected, since it costs less in PW terms; that is, the PW B value is numerically
larger than PW A .
(b) For a 5-year study period, no cycle repeats are necessary. The PW analysis is
PW A 15,000 3500(PA,15%,5) 1000(PF,15%,5)
$26,236
PW B 18,000 3100(PA,15%,5) 2000(PF,15%,5)
$27,397
Vendor A is now selected based on its smaller PW value. This means that the shortened
study period of 5 years has caused a switch in the economic decision. In situations such as
this, the standard practice of using a fixed study period should be carefully examined to
ensure that the appropriate approach, that is, LCM or fixed study period, is used to satisfy
the equal-service requirement.
EXAMPLE 5.4 Water for Semiconductor Manufacturing Case
PE
When we discussed this case in the introduction, we learned that the initial estimates of equipment
life were 10 years for both options of UPW (ultrapure water)—seawater and groundwater.
As you might guess, a little research indicates that seawater is more corrosive and the equipment
life is shorter—5 years rather than 10. However, it is expected that, instead of complete replacement,
a total refurbishment of the equipment for $10 M after 5 years will extend the life through
the anticipated 10 th year of service.
With all other estimates remaining the same, it is important to determine if this 50%
reduction in expected usable life and the refurbishment expense may alter the decision to go
with the seawater option, as determined in Example 5.2. For a complete analysis, consider
both a 10-year and a 5-year option for the expected use of the equipment, regardless of the
source of UPW.

136 Chapter 5 Present Worth Analysis
Seawater option
OC UPW cost/year
0.5 1.44
Groundwater option
OC UPW cost/year
0.3 1.80
Both options
Salvage values
included here
Salvage values are the
same after 5 and 10 years
Figure 5–3
PW analyses using LCM and study period approaches for water for semiconductor manufacturing case,
Example 5.4.
Solution
A spreadsheet and the NPV function are a quick and easy way to perform this dual analysis.
The details are presented in Figure 5–3.
LCM of 10 years: In the top part of the spreadsheet, the LCM of 10 years is necessary to satisfy
the equal-service requirement; however, the first cost in year 5 is the refurbishment cost of
$10 M, not the $20 M expended in year 0. Each year’s cash flow is entered in consecutive
cells; the $11.94 M in year 5 accounts for the continuing AOC and annual UPW cost of
$1.94 M, plus the $10 M refurbishment cost. The NPV functions shown on the spreadsheet
determine the 12% per year PW values in $1 million units.
PW S $36.31
PW G $33.16
Now, the groundwater option is cheaper ; the economic decision is reversed with this new
e stimate of life and year 5 refurbishment expense.
Study period of 5 years: The lower portion of Figure 5–3 details a PW analysis using the
second approach to evaluating different-life alternatives, that is, a specific study period, which
is 5 years in this case study. Therefore, all cash flows after 5 years are neglected.
Again the economic decision is reversed as the 12% per year PW values favor the seawater
option.
PW S $26.43
PW G $28.32
Comments
The decision switched between the LCM and study period approaches. Both are correct answers
given the decision of how the equal-service requirement is met. This analysis demonstrates
how important it is to compare mutually exclusive alternatives over time periods that
are believable and to take the time necessary to make the most accurate cost, life, and MARR
estimates when the evaluation is performed.

5.4 Future Worth Analysis 137
If the PW evaluation is incorrectly performed using the respective lives of the two options,
the equal-service requirement is violated, and PW values favor the shorter-lived option, that is,
seawater. The PW values are
Option S: n 5 years, PW S $26.43 M, from the bottom left calculation in Figure 5–3.
Option G: n 10 years, PW G $33.16 M, from the top right calculation in Figure 5–3.
For independent projects , use of the LCM approach is unnecessary since each project is compared
to the do-nothing alternative, not to each other, and satisfying the equal-service requirement
is not a problem. Simply use the MARR to determine the PW over the respective life of
each project, and select all projects with a PW 0.
5.4 Future Worth Analysis
The future worth (FW) of an alternative may be determined directly from the cash flows, or by
multiplying the PW value by the F P factor, at the established MARR. The n value in the F P
factor is either the LCM value or a specified study period. Analysis of alternatives using FW
values is especially applicable to large capital investment decisions when a prime goal is to
maximize the future wealth of a corporation’s stockholders.
Future worth analysis over a specified study period is often utilized if the asset (equipment, a
building, etc.) might be sold or traded at some time before the expected life is reached. Suppose
an entrepreneur is planning to buy a company and expects to trade it within 3 years. FW analysis
is the best method to help with the decision to sell or keep it 3 years hence. Example 5.5 illustrates
this use of FW analysis. Another excellent application of FW analysis is for projects that will come
online at the end of a multiyear investment period, such as electric generation facilities, toll roads,
airports, and the like. They are analyzed using the FW value of investment commitments made
during construction.
The selection guidelines for FW analysis are the same as for PW analysis; FW 0 means the
MARR is met or exceeded. For two or more mutually exclusive alternatives, select the one
with the numerically largest FW value.
ME alternative
selection
EXAMPLE 5.5
A British food distribution conglomerate purchased a Canadian food store chain for £75 million
3 years ago. There was a net loss of £10 million at the end of year 1 of ownership. Net cash
flow is increasing with an arithmetic gradient of £5 million per year starting the second year,
and this pattern is expected to continue for the foreseeable future. This means that breakeven
net cash flow was achieved this year. Because of the heavy debt financing used to purchase the
Canadian chain, the international board of directors expects a MARR of 25% per year from
any sale.
(a) The British conglomerate has just been offered £159.5 million by a French company
wishing to get a foothold in Canada. Use FW analysis to determine if the MARR will be
realized at this selling price.
(b) If the British conglomerate continues to own the chain, what selling price must be obtained
at the end of 5 years of ownership to just make the MARR?
Solution
(a) Set up the future worth relation in year 3 (FW 3 ) at i 25% per year and an offer price of
£159.5 million. Figure 5–4 a presents the cash flow diagram in million £ units.
FW 3 75(FP,25%,3) 10(FP,25%,2) 5(FP,25%,1) 159.5
168.36 159.5 £8.86 million
No, the MARR of 25% will not be realized if the £159.5 million offer is accepted.

138 Chapter 5 Present Worth Analysis
£159.5
FW = ?
i 25%?
i = 25%
£5
£10
0 1 2 3
Year
0 1 2 3 4 5
Year
£10
£5
£10
£5
£75
(a)
£75
Figure 5–4
Cash flow diagrams for Example 5.5. (a) i ?; (b) FW ?.
(b)
(b) Determine the future worth 5 years from now at 25% per year. Figure 5–4 b presents the
cash flow diagram. The A G and F A factors are applied to the arithmetic gradient.
FW 5 75(FP,25%,5) 10(FA,25%,5) 5(AG,25%,5)(FA,25%,5)
£246.81 million
The offer must be for at least £246.81 million to make the MARR. This is approximately
3.3 times the purchase price only 5 years earlier, in large part based on the required
MARR of 25%.
5.5 Capitalized Cost Analysis
Many public sector projects such as bridges, dams, highways and toll roads, railroads, and hydroelectric
and other power generation facilities have very long expected useful lives. A perpetual
or infinite life is the effective planning horizon. Permanent endowments for charitable organizations
and universities also have perpetual lives. The economic worth of these types of projects or
endowments is evaluated using the present worth of the cash flows.
Capitalized Cost (CC) is the present worth of a project that has a very long life (more than, say,
35 or 40 years) or when the planning horizon is considered very long or infinite.
The formula to calculate CC is derived from the PW relation P A ( P A , i %, n ), where n
time periods. Take the equation for P using the P A factor and divide the numerator and denominator
by (1 i ) to obtain
n
P A
[ 1 ———— 1
(1 i) n
————— ]
i
As n approaches , the bracketed term becomes 1 i . We replace the symbols P and PW with
CC as a reminder that this is a capitalized cost equivalence. Since the A value can also be termed
AW for annual worth, the capitalized cost formula is simply
CC A —
i
or
CC AW ——
i
[5.1]
Solving for A or AW, the amount of new money that is generated each year by a capitalization
of an amount CC is
AW CC (i) [5.2]
This is the same as the calculation A P (i ) for an infinite number of time periods. Equation [5.2]
can be explained by considering the time value of money. If $20,000 is invested now (this is the

5.5 Capitalized Cost Analysis 139
capitalization) at 10% per year, the maximum amount of money that can be withdrawn at the end
of every year for eternity is $2000, which is the interest accumulated each year. This leaves the
original $20,000 to earn interest so that another $2000 will be accumulated the next year.
The cash flows (costs, revenues, and savings) in a capitalized cost calculation are usually of
two types: recurring, also called periodic, and nonrecurring. An annual operating cost of $50,000
and a rework cost estimated at $40,000 every 12 years are examples of recurring cash flows.
Examples of nonrecurring cash flows are the initial investment amount in year 0 and one-time
cash flow estimates at future times, for example, $500,000 in fees 2 years hence.
The procedure to determine the CC for an infinite sequence of cash flows is as follows:
1. Draw a cash flow diagram showing all nonrecurring (one-time) cash flows and at least two
cycles of all recurring (periodic) cash flows.
2. Find the present worth of all nonrecurring amounts. This is their CC value.
3. Find the A value through one life cycle of all recurring amounts. (This is the same value in
all succeeding life cycles, as explained in Chapter 6.) Add this to all other uniform
amounts (A) occurring in years 1 through infinity. The result is the total equivalent uniform
annual worth (AW).
4. Divide the AW obtained in step 3 by the interest rate i to obtain a CC value. This is an
application of Equation [5.1].
5. Add the CC values obtained in steps 2 and 4.
Drawing the cash flow diagram (step 1) is more important in CC calculations than elsewhere,
because it helps separate nonrecurring and recurring amounts. In step 5 the present worths of all
component cash flows have been obtained; the total capitalized cost is simply their sum.
EXAMPLE 5.6
The Haverty County Transportation Authority (HCTA) has just installed new software to charge
and track toll fees. The director wants to know the total equivalent cost of all future costs incurred
to purchase the software system. If the new system will be used for the indefinite future,
find the equivalent cost ( a ) now, a CC value, and ( b ) for each year hereafter, an AW value.
The system has an installed cost of $150,000 and an additional cost of $50,000 after
10 years. The annual software maintenance contract cost is $5000 for the first 4 years and
$8000 thereafter. In addition, there is expected to be a recurring major upgrade cost of $15,000
every 13 years. Assume that i 5% per year for county funds.
Solution
(a) The five-step procedure to find CC now is applied.
1. Draw a cash flow diagram for two cycles (Figure 5–5).
2. Find the present worth of the nonrecurring costs of $150,000 now and $50,000 in year
10 at i 5%. Label this CC 1 .
CC 1 150,000 50,000(PF,5%,10) $180,695
0 2 4 6 8 10 12 14 20 26
Year
$5000
$8000
i = 5% per year
$50,000
$15,000 $15,000
$150,000
Figure 5–5
Cash flows for two cycles of recurring costs and all nonrecurring amounts, Example 5.6.

140 Chapter 5 Present Worth Analysis
3 and 4. Convert the $15,000 recurring cost to an A value over the first cycle of 13 years, and
find the capitalized cost CC 2 at 5% per year using Equation [5.1].
A 15,000(AF,5%,13) $847
CC 2 8470.05 $16,940
There are several ways to convert the annual software maintenance cost series to A
and CC values. A straightforward method is to, first, consider the $−5000 an A series
with a capitalized cost of
CC 3 50000.05 $100,000
Second, convert the step-up maintenance cost series of $−3000 to a capitalized cost
CC 4 in year 4, and find the present worth in year 0. (Refer to Figure 5–5 for cash flow
timings.)
CC 4 ————
3,000 (PF,5%,4) $49,362
0.05
5. The total capitalized cost CC T for Haverty County Transportation Authority is the sum
of the four component CC values.
CC T 180,695 16,940 100,000 49,362
$346,997
(b) Equation [5.2] determines the AW value forever.
AW Pi CC T (i) $346,997(0.05) $17,350
Correctly interpreted, this means Haverty County officials have committed the equivalent
of $17,350 forever to operate and maintain the toll management software.
For the comparison of two alternatives on the basis of capitalized cost, use the procedure
above to find the A value and CC T for each alternative. Since the capitalized cost represents the
total present worth of financing and maintaining a given alternative forever, the alternatives will
automatically be compared for the same number of years (i.e., infinity). The alternative with the
smaller capitalized cost will represent the more economical one. This evaluation is illustrated in
Example 5.7 using the progressive example for this chapter.
EXAMPLE 5.7 Water for Semiconductor Manufacturing Case
Our case study has progressed (in Example 5.4) to the point that the life of the seawater option
can be extended to 10 years with a major refurbishment cost after 5 years. This extension is
possible only one time, after which a new life cycle would commence. In $1 million units, the
estimates and PW values (from Figure 5–3) are as follows:
Seawater: P S $20; AOC S $1.94; n S 10 years; refurbishment, year 5 $10;
S S 0.05(20) $1.00; PW S $36.31
Groundwater: P G $22; AOC G $2.10; n G 10 years; S G 0.10(22) $2.2;
PW G $33.16
PE
If we assume that the UPW (ultrapure water) requirement will continue for the foreseeable
future, a good number to know is the present worth of the long-term options at the selected
MARR of 12% per year. What are these capitalized costs for the two options using the estimates
made thus far?
Solution
Find the equivalent A value for each option over its respective life, then determine the CC
value using the relation CC A i . Select the option with the lower CC. This approach satisfies
the equal-service requirement because the time horizon is infinity when the CC is determined.

5.5 Capitalized Cost Analysis 141
Seawater: A S PW S (AP,12%,10) 36.31(0.17698) $6.43
CC S 6.430.12 $53.58
Groundwater: A G PW G (AP,12%,10) 33.16(0.17698) $5.87
CC G 5.870.12 $48.91
In terms of capitalized cost, the groundwater alternative is cheaper.
Comment
If the seawater-life extension is not considered a viable option, the original alternative of
5 years could be used in this analysis. In this case, the equivalent A value and CC computations
in $1 million units are as follows.
A S,5 years 20(AP,12%,5) 1.94 0.05(20)(AF,12%,5)
$7.33
CC S, 5 years 7.33/0.12 $61.08
Now, the economic advantage of the groundwater option is even larger.
If a finite-life alternative (for example, 5 years) is compared to one with an indefinite or very
long life, capitalized costs can be used. To determine capitalized cost for the finite life alternative,
calculate the equivalent A value for one life cycle and divide by the interest rate (Equation [5.1]).
This procedure is illustrated in Example 5.8 using a spreadsheet.
EXAMPLE 5.8
The State Legislature has mandated a statewide recycling program to include all types of
plastic, paper, metal, and glass refuse. The goal is zero landfill by 2020. Two options for the
materials separation equipment are outlined below. The interest rate for state-mandated projects
is 5% per year.
Contractor option (C): $8 million now and $25,000 per year will provide separation services
at a maximum of 15 sites. No contract period is stated; thus the contract and services are offered
for as long as the State needs them.
Purchase option (P): Purchase equipment at each site for $275,000 per site and expend an
estimated $12,000 in annual operating costs (AOC). Expected life of the equipment is 5 years
with no salvage value.
(a) Perform a capitalized cost analysis for a total of 10 recycling sites.
(b) Determine the maximum number of sites at which the equipment can be purchased and
still have a capitalized cost less than that of the contractor option.
Solution
(a) Figure 5–6, column B, details the solution. The contract, as proposed, has a long life.
Therefore, the $8 million is already a capitalized cost. The annual charge of A $25,000
is divided by i 0.05 to determine its CC value. Summing the two values results in
CC C $8.5 million.
For the finite, 5-year purchase alternative, column B shows the first cost ($275,000 per
site), AOC ($12,000), and equivalent A value of $−755,181, which is determined via
the PMT function (cell tag). Divide A by the interest rate of 5% to determine CC P
$15.1 million.
The contractor option is by far more economical for the anticipated 10 sites.
(b) A quick way to find the maximum number of sites for which CC P CC C is to use Excel’s
Goal Seek tool, introduced in Chapter 2, Example 2.10. (See Appendix A for details
on how to use this tool.) The template is set up in Figure 5–6 to make the two CC values
equal as the number of sites is altered (decreased). The result, shown in column C, indicates
that 5.63 sites make the options economically equivalent. Since the number of sites

142 Chapter 5 Present Worth Analysis
PMT($B$1,B13,B12*B4) B14*B4
Figure 5–6
Spreadsheet solution of Example 5.8 using capitalized cost ( a ) for 10 recycling sites and ( b ) to determine the
number of sites to make the alternatives economically equal.
must be an integer, 5 or fewer sites will favor purchasing the equipment and 6 or more
sites will favor contracting the separation services.
This approach to problem solution will be called breakeven analysis in later chapters
of the text. By the way, another way to determine the number of sites is by trial
and error. Enter different values in cell B4 until the CC values favor the purchase
alternative.
CHAPTER SUMMARY
The present worth method of comparing alternatives involves converting all cash flows to present
dollars at the MARR. The alternative with the numerically larger (or largest) PW value is selected.
When the alternatives have different lives, the comparison must be made for equal-service
periods. This is done by performing the comparison over either the LCM of lives or a specific
study period. Both approaches compare alternatives in accordance with the equal-service requirement.
When a study period is used, any remaining value in an alternative is recognized
through the estimated future market value.
If the life of the alternatives is considered to be very long or infinite, capitalized cost is the
comparison method. The CC value is calculated as A i , because the P A factor reduces to 1 i in
the limit of n .
PROBLEMS
Types of Projects
5.1 What is the difference between mutually exclusive
alternatives and independent projects?
5.2 (a) What is meant by the do-nothing alternative?
(b) When is the do-nothing alternative not an
option?
5.3 (a) How many alternatives are possible from
four independent projects identified as W, X,
Y, and Z?
(b) List all of the possibilities.
5.4 What is the difference between a revenue and a
cost alternative?

Problems 143
5.5 What is meant by the term equal service ?
5.6 What two approaches can be used to satisfy the
equal-service requirement?
Alternative Comparison—Equal Lives
5.7 A company that manufactures magnetic membrane
switches is investigating two production options
that have the estimated cash flows shown ($1 million
units). Which one should be selected on the
basis of a present worth analysis at 10% per year?
In-house
Contract
First cost, $ 30 0
Annual cost, $ per year 5 2
Annual income, $ per year 14 3.1
Salvage value, $ 2 —
Life, years 5 5
5.8 The manager of a canned food processing plant
must decide between two different labeling machines.
Machine A will have a first cost of $42,000,
an annual operating cost of $28,000, and a service
life of 4 years. Machine B will cost $51,000 to buy
and will have an annual operating cost of $17,000
during its 4-year life. At an interest rate of 10% per
year, which should be selected on the basis of a
present worth analysis?
5.9 A metallurgical engineer is considering two materials
for use in a space vehicle. All estimates are
made. ( a ) Which should be selected on the basis of
a present worth comparison at an interest rate of
12% per year? ( b ) At what first cost for the material
not selected above will it become the more
economic alternative?
Material X
Material Y
First cost, $ 15,000 35,000
Maintenance cost, $ per year 9,000 7,000
Salvage value, $ 2,000 20,000
Life, years 5 5
5.10 To retain high-performing engineers, a large semiconductor
company provides corporate stock as
part of the compensation package. In one particular
year, the company offered 1000 shares of either
class A or class B stock. The class A stock was selling
for $30 per share at the time, and stock market
analysts predicted that it would increase at a rate of
6% per year for the next 5 years. Class B stock was
selling for $20 per share, but its price was expected
to increase by 12% per year. At an interest rate of
8% per year, which stock should the engineers select
on the basis of a present worth analysis and a
5-year planning horizon?
5.11 The Murphy County Fire Department is considering
two options for upgrading its aging physical facilities.
Plan A involves remodeling the fire stations
on Alameda Avenue and Trowbridge Boulevard that
are 57 and 61 years old, respectively. (The industry
standard is about 50 years of use for a station.) The
cost for remodeling the Alameda station is estimated
at $952,000 while the cost of redoing the
Trowbridge station is $1.3 million. Plan B calls for
buying 5 acres of land somewhere between the two
stations, building a new fire station, and selling the
land and structures at the previous sites. The cost of
land in that area is estimated to be $366,000 per
acre. The size of the new fire station would be 9000
square feet with a construction cost of $151.18 per
square foot. Contractor fees for overhead, profit,
etc. are expected to be $340,000, and architect fees
will be $81,500. (Assume all of the costs for plan B
occur at time 0.) If plan A is adopted, the extra cost
for personnel and equipment will be $126,000 per
year. Under plan B, the sale of the old sites is anticipated
to net a positive $500,000 five years in the
future. Use an interest rate of 6% per year and a 50-
year useful life for the remodeled and new stations
to determine which plan is better on the basis of a
present worth analysis.
5.12 Delcon Properties is a commercial developer of
shopping centers and malls in various places around
the country. The company needs to analyze the economic
feasibility of rainwater drains in a 60-acre
area that it plans to develop. Since the development
won’t be started for 3 years, this large open space
will be subject to damage from heavy thunderstorms
that cause soil erosion and heavy rutting. If
no drains are installed, the cost of refilling and
grading the washed out area is expected to be
$1500 per thunderstorm. Alternatively, a temporary
corrugated steel drainage pipe could be installed
that will prevent the soil erosion. The cost of the
pipe will be $3 per foot for the total length of
7000 feet required. Some of the pipe will be salvageable
for $4000 at the end of the 3-year period
between now and when the construction begins.
Assuming that thunderstorms occur regularly at
3-month intervals, starting 3 months from now,
which alternative should be selected on the basis of
a present worth comparison using an interest rate of
4% per quarter?
5.13 A public water utility is trying to decide between
two different sizes of pipe for a new water main. A
250-mm line will have an initial cost of $155,000,
whereas a 300-mm line will cost $210,000. Since
there is more head loss through the 250-mm pipe,
the pumping cost is expected to be $3000 more per
year than for the 250-mm line. If the lines are

144 Chapter 5 Present Worth Analysis
expected to last for 30 years, which size should be
selected on the basis of a present worth analysis
using an interest rate of 10% per year?
5.14 The supervisor of a community swimming pool
has developed two methods for chlorinating the
pool. If gaseous chlorine is added, a chlorinator
will be required that has an initial cost of $8000
and a useful life of 5 years. The chlorine will cost
$650 per year, and the labor cost will be $800 per
year. Alternatively, dry chlorine can be added manually
at a cost of $1000 per year for chlorine and
$1900 per year for labor. Which method should be
used on the basis of a present worth analysis if the
interest rate is 10% per year?
5.15 Anion, an environmental engineering consulting
firm, is trying to be eco-friendly in acquiring an automobile
for general office use. It is considering a
gasoline-electric hybrid and a gasoline-free allelectric
hatchback. The hybrid under consideration
is GM’s Volt, which will cost $35,000 and have a
range of 40 miles on the electric battery and several
hundred more miles when the gasoline engine
kicks in. Nissan’s Leaf, on the other hand, is a pure
electric that will have a range of only 100 miles,
after which its lithium-ion battery will have to be
recharged. The Leaf’s relatively limited range creates
a psychological effect known as range anxiety.
This fact alone has caused the company to lean toward
purchasing the Volt, which is assumed to have
a salvage value of $15,000 in 5 years. The Leaf
could be leased for $349 per month (end-of-month
payments) for 5 years after an initial $1500 down
payment for “account activation.” If the consulting
company plans to ignore the range anxiety effect in
making its decision, which automobile is the better
option on the basis of a present worth analysis at an
interest rate of 0.75% per month? Assume the operating
cost will be the same for both vehicles.
5.16 A pipeline engineer working in Kuwait for the oil
giant BP wants to perform a present worth analysis on
alternative pipeline routings—the first predominately
by land and the second primarily undersea. The undersea
route is more expensive initially due to extra
corrosion protection and installation costs, but cheaper
security and maintenance reduces annual costs. Perform
the analysis for the engineer at 15% per year.
Land
Undersea
Installation cost, $ million 215 350
Pumping, operating, security, 22 2
$ million per year
Replacement of valves and
30 70
appurtenances in year 25, $ million
Expected life, years 50 50
Alternative Comparison—Different Lives
5.17 An electric switch manufacturing company has to
choose one of three different assembly methods.
Method A will have a first cost of $40,000, an annual
operating cost of $9000, and a service life of
2 years. Method B will cost $80,000 to buy and
will have an annual operating cost of $6000 over
its 4-year service life. Method C will cost $130,000
initially with an annual operating cost of $4000
over its 8-year life. Methods A and B will have no
salvage value, but method C will have some equipment
worth an estimated $12,000. Which method
should be selected? Use present worth analysis at
an interest rate of 10% per year.
5.18 Midwest Power and Light operates 14 coal-fired
power plants in several states around the United
States. The company recently settled a lawsuit by
agreeing to pay $60 million in mitigation costs related
to acid rain. The settlement included $21 million
to reduce emissions from barges and trucks in
the Ohio River Valley, $24 million for projects to
conserve energy and produce alternative energy,
$3 million for Chesapeake Bay, $2 million for
Shenandoah National Park, and $10 million to acquire
ecologically sensitive lands in Appalachia.
The question of how to distribute the money over
time has been posed. Plan A involves spending
$5 million now and the remaining $55 million
equally over a 10-year period (that is, $5.5 million
in each of years 1 through 10). Plan B requires expenditures
of $5 million now, $25 million 2 years
from now, and $30 million 7 years from now. Determine
which plan is more economical on the
basis of a present worth analysis over a 10-year
period at an interest rate of 10% per year.
5.19 Machines that have the following costs are under
consideration for a robotized welding process.
Using an interest rate of 10% per year, determine
which alternative should be selected on the basis
of a present worth analysis. Show ( a ) hand and
( b ) spreadsheet solutions.
Machine X Machine Y
First cost, $ 250,000 430,000
Annual operating cost, $ per year 60,000 40,000
Salvage value, $ 70,000 95,000
Life, years 3 6
5.20 Water for Semiconductor Manufacturing Case
PE
Throughout the present worth analyses, the decision
between seawater and groundwater switched
multiple times in Examples 5.2 and 5.4. A summary
is given here in $1 million units.

Life n ,
years
First
cost, $
Seawater (S)
PW at
12%, $ Selected
Life n ,
years
Problems 145
Groundwater (G)
First
cost, $
PW at
12%, $ Selected
10 20 30.64 Yes 10 22 33.16 No
5
(study
period)
5 20, plus
10 after
5 years
36.31 No 10 22 33.16 Yes
20 26.43 Yes 5
(study
period)
22 28.32 No
The confusion about the recommended source for
UPW has not gone unnoticed by the general manager.
Yesterday, you were asked to settle the issue
by determining the first cost X S of the seawater option
to ensure that it is the economic choice over
groundwater. The study period is set by the manager
as 10 years, simply because that is the time period on
the lease agreement for the building where the fab
will be located. Since the seawater equipment must
be refurbished or replaced after 5 years, the general
manager told you to assume that the equipment will
be purchased anew after 5 years of use. What is the
maximum first cost that Angular Enterprises should
pay for the seawater option?
5.21 Accurate airflow measurement requires straight
unobstructed pipe for a minimum of 10 diameters
upstream and 5 diameters downstream of the measuring
device. In a field application, physical constraints
compromise the pipe layout, so the engineer
is considering installing the airflow probes in an
elbow, knowing that flow measurement will be less
accurate but good enough for process control. This
is plan 1, which will be in place for only 3 years,
after which a more accurate flow measurement system
with the same costs as plan 1 will be available.
This plan will have a first cost of $26,000 with an
annual maintenance cost estimated at $5000.
Plan 2 involves installation of a recently designed
submersible airflow probe. The stainless
steel probe can be installed in a drop pipe with the
transmitter located in a waterproof enclosure on the
handrail. The first cost of this system is $83,000,
but because it is accurate and more durable, it will
not have to be replaced for at least 6 years. Its maintenance
cost is estimated to be $1400 per year plus
$2500 in year 3 for replacement of signal processing
software. Neither system will have a salvage
value. At an interest rate of 10% per year, which
one should be selected on the basis of a present
worth comparison?
5.22 An engineer is considering two different liners for an
evaporation pond that will receive salty concentrate
from a brackish water desalting plant. A plastic liner
will cost $0.90 per square foot initially and will
require replacement in 15 years when precipitated
solids will have to be removed from the pond using
heavy equipment. This removal will cost $500,000.
A rubberized elastomeric liner is tougher and, therefore,
is expected to last 30 years, but it will cost
$2.20 per square foot. If the size of the pond is
110 acres (1 acre 43,560 square feet), which liner
is more cost effective on the basis of a present worth
comparison at an interest rate of 8% per year?
5.23 A sports mortgage is the brainchild of Stadium Capital
Financing Group, a company headquartered in
Chicago, Illinois. It is an innovative way to finance
cash-strapped sports programs by allowing fans to
sign up to pay a “mortgage” over a certain number of
years for the right to buy good seats at football games
for several decades with season ticket prices locked
in at current prices. In California, the locked-in price
period is 50 years. Assume UCLA fan X purchases a
$130,000 mortgage and pays for it now to get season
tickets for $290 each for 50 years, while fan Y buys
season tickets at $290 in year 1, with prices increasing
by $20 per year for 50 years. ( a ) Which fan made
the better deal if the interest rate is 8% per year?
( b ) What should fan X be willing to pay up front for
the mortgage to make the two plans exactly equivalent
economically? (Assume he has no reason to
give extra money to UCLA at this point.)
5.24 A chemical processing corporation is considering
three methods to dispose of a non-hazardous chemical
sludge: land application, fluidized-bed incineration,
and private disposal contract. The estimates for
each method are shown. Determine which has the
least cost on the basis of a present worth comparison
at 10% per year for the following scenarios:
(a) The estimates as shown
(b) The contract award cost increases by 20%
every 2-year renewal
Land Application Incineration
Contract
First cost, $ 130,000 900,000 0
Annual operating 95,000 60,000 120,000
cost, $ per year
Salvage value, $ 25,000 300,000 0
Life, years 3 6 2

146 Chapter 5 Present Worth Analysis
5.25 An assistant to Stacy gave her the PW values for
four alternatives they are comparing for the development
of a remote control vibration control system
for offshore platform application. The results
in the table use a MARR of 14% per year. Determine
which alternative(s) should be selected ( a ) if
the alternatives are exclusive, and ( b ) if the projects
are independent.
I J K L
Life n , years 3 4 12 6
PW over n years, $ 16.08 31.12 257.46 140.46
PW over 6 years, $ 26.94 15.78 653.29 140.46
PW over 12 years, $ 39.21 60.45 257.46 204.46
Future Worth Comparison
5.26 An industrial engineer is considering two robots
for purchase by a fiber-optic manufacturing company.
Robot X will have a first cost of $80,000, an
annual maintenance and operation (M&O) cost of
$30,000, and a $40,000 salvage value. Robot Y
will have a first cost of $97,000, an annual M&O
cost of $27,000, and a $50,000 salvage value.
Which should be selected on the basis of a future
worth comparison at an interest rate of 15% per
year? Use a 3-year study period.
5.27 Two processes can be used for producing a polymer
that reduces friction loss in engines. Process T
will have a first cost of $750,000, an operating cost
of $60,000 per year, and a salvage value of $80,000
after its 2-year life. Process W will have a first cost
of $1,350,000, an operating cost of $25,000 per
year, and a $120,000 salvage value after its 4-year
life. Process W will also require updating at the
end of year 2 at a cost of $90,000. Which process
should be selected on the basis of a future worth
analysis at an interest rate of 12% per year?
5.28 Compare the alternatives shown below on the
basis of a future worth analysis, using an interest
rate of 8% per year.
First cost, $ 23,000 30,000
Annual operating cost, $ per year 4,000 2,500
Salvage value, $ 3,000 1,000
Life, years 3 6
5.29 Two manufacturers supply MRI systems for medical
imaging. St. Jude’s Hospital wishes to replace its
current MRI equipment that was purchased 8 years
ago with the newer technology and clarity of a stateof-the-art
system. System K will have a first cost of
$1,600,000, an operating cost of $70,000 per year,
and a salvage value of $400,000 after its 4-year life.
System L will have a first cost of $2,100,000, an
P
Q
operating cost of $50,000 the first year with an expected
increase of $3000 per year thereafter, and no
salvage value after its 8-year life. Which system
should be selected on the basis of a future worth
analysis at an interest rate of 12% per year?
5.30 A retail shopping center developer signed a contract
to build a $100 million high-end shopping center in
City Center, because the city and county governments
agreed to sales and tax rebates totaling
$18.7 million over 10 years. The contract called for
the developer to raze existing buildings 2 years
from the date the contract was signed and to have
the shopping center built by the end of year 3. However,
due to a real estate–induced recession in the
United States, the developer sought and was granted
a new contract. The new contract required the developer
to raze the existing buildings at the end of
year 1, but the shopping center would not have to be
completed for 7 years from the date the contract was
signed. Assume that the cost for razing the existing
buildings is $1.3 million and the developer does not
build the shopping center until 7 years from now (at
a cost of $100 million). Determine the difference in
the future worth cost in year 7 of the two contracts
at an interest rate of 10% per year.
Capitalized Cost
5.31 A wealthy businessman wants to start a permanent
fund for supporting research directed toward sustainability.
The donor plans to give equal amounts
of money for each of the next 5 years, plus one now
(i.e., six donations) so that $100,000 per year can
be withdrawn each year forever, beginning in
year 6. If the fund earns interest at a rate of 8% per
year, how much money must be donated each time?
5.32 Bob, a philanthropist, is not sure what rate of return
his gifts may realize once donated to his favorite
charity. Determine the capitalized cost of $10,000
every 5 years forever , starting 5 years from now
at an interest rate of ( a ) 3% and ( b ) 8% per year.
( c ) Explain the significant difference between the
two capitalized costs.
5.33 Find the capitalized cost of a present cost of
$300,000, annual costs of $35,000, and periodic
costs every 5 years of $75,000. Use an interest rate
of 12% per year.
5.34 Water for Semiconductor Manufacturing Case
PE
It is anticipated that the needs for UPW (ultrapure
water) at the new Angular Enterprises site will
continue for a long time, as long as 50 years. This
is the rationale for using capitalized cost as a basis
for the economic decision between desalinated

Additional Problems and FE Exam Review Questions 147
seawater (S) and purified groundwater (G). These
costs were determined (Example 5.7) to be CC S
$53.58 million and CC G $48.91 million.
Groundwater is the clear economic choice.
Yesterday, the general manager had lunch with
the president of Brissa Water, who offered to supply
the needed UPW at a cost of $5 million per year
for the indefinite future. It would mean a dependence
upon a contractor to supply the water, but the
equipment, treatment, and other costly activities to
obtain UPW on-site would be eliminated. The
manager asks you to make a recommendation
about this seemingly attractive alternative under
the following conditions at the same MARR of
12% per year as used for the other analyses:
( a) The annual cost of $5 million remains constant
throughout the time it is needed.
(b) The annual cost starts at $5 million for the first
year only, and then it increases 2% per year.
(This increase is above the cost of providing
UPW by either of the other two methods.)
5.35 Compare the alternatives shown on the basis of
their capitalized costs using an interest rate of
10% per year.
Alternative M
Alternative N
First cost, $ 150,000 800,000
Annual operating 50,000 12,000
cost, $ per year
Salvage value, $ 8,000 1,000,000
Life, years 5
5.36 The cost of maintaining a certain permanent monument
in Washington, DC occurs as periodic outlays
of $1000 every year and $5000 every 4 years.
Calculate the capitalized cost of the maintenance
using an interest rate of 10% per year.
5.37 Because you are thankful for what you learned in
engineering economy, you plan to start a permanent
scholarship fund in the name of the professor
who taught the course. You plan to deposit money
now with the stipulation that the scholarships be
awarded beginning 12 years from now (which
happens to be the exact time that your daughter
plans to begin college). The interest that is accumulated
between now and year 12 is to be added to
the principal of the endowment. After that, the interest
that is earned each year will be awarded as
scholarship money. If you want the amount of the
scholarships to be $40,000 per year, how much
must you donate now if the fund earns interest at a
rate of 8% per year?
5.38 A patriotic group of firefighters is raising money
to erect a permanent (i.e., infinite life) monument
in New York City to honor those killed in the line
of duty. The initial cost of the monument will be
$150,000, and the annual maintenance will cost
$5000. There will be an additional one-time cost
of $20,000 in 2 years to add names of those who
were missed initially. At an interest rate of 6% per
year, how much money must they raise now in
order to construct and maintain the monument
forever?
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
5.39 One assumption inherent in the present worth
method of analysis is that:
( a) The alternatives will be used only through
the life of the shortest-lived alternative.
( b) The alternatives will be used only through
the life of the longest-lived alternative.
( c) The cash flows of each alternative will
change only by the inflation or deflation rate
in succeeding life cycles.
( d) At least one of the alternatives will have a
finite life.
5.40 When only one alternative can be selected from
two or more, the alternatives are said to be:
( a) Mutually exclusive
( b) Independent alternatives
( c) Cost alternatives
( d) Revenue alternatives
5.41 For the mutually exclusive alternatives shown, the
one(s) that should be selected are:
(a)
(b)
(c)
(d)
Alternative PW, $
Only C
Only A
C and D
Only D
A 25,000
B 12,000
C 10,000
D 15,000
5.42 The alternatives shown are to be compared on the
basis of their present worth values. At an interest
rate of 10% per year, the values of n that you
should use in the uniform series factors to make a

148 Chapter 5 Present Worth Analysis
correct comparison by the present worth method
are:
Alternative A
Alternative B
First cost, $ 50,000 90,000
Annual operating cost, 10,000 4000
$ per year
Salvage value, $ 13,000 15,000
Life, years 3 6
(a)
(b)
(c)
(d)
n 3 years for A and n 3 years for B
n 3 years for A and n 6 years for B
n 6 years for A and n 6 years for B
None of the above
5.43 The value of the future worth for alternative P at an
interest rate of 8% per year is closest to:
P
Q
First cost, $ 23,000 30,000
Annual operating cost, 4,000 2,500
$ per year
Salvage value, $ 3,000 1,000
Life, years 3 6
(a)
(b)
(c)
(d)
FW P $88,036
FW P $86,026
FW P $81,274
FW P $70,178
5.44 The present worth of $50,000 now, $10,000 per year
in years 1 through 15, and $20,000 per year in years
16 through infinity at 10% per year is closest to:
(a) Less than $169,000
(b) $169,580
(c) $173,940
(d) $195,730
5.45 A donor (you) wishes to start an endowment that
will provide scholarship money of $40,000 per year
beginning in year 5 and continuing indefinitely. If
the university earns 10% per year on the endowment,
the amount you must donate now is closest to:
(a) $225,470
(b) $248,360
(c) $273,200
(d) $293,820
Problems 5.46 through 5.48 are based on the following
information.
Alternative I
Alternative J
Initial cost, $ 150,000 250,000
Annual income, $ per year 20,000 40,000
Annual expenses, $ per year 9,000 14,000
Salvage value, $ 25,000 35,000
Life, years 3 6
The interest rate is 15% per year.
5.46 In comparing alternatives I and J by the present
worth method, the value of n that must be used in
11,000( P A , i , n ) for alternative I is:
(a) 3
(b) 6
(c) 18
(d) 36
5.47 In comparing alternatives I and J by the present
worth method, the equation that yields the present
worth of alternative J is:
(a) PW J 250,000 40,000( P A ,15%,6)
35,000( P F ,15%,6)
(b) PW J 250,000 26,000( P A ,15%,6)
( c) PW J
35,000( P F ,15%,6)
250,000 26,000( P A ,15%,6)
35,000( P F ,15%,6)
( d) PW J 250,000 26,000( P A ,15%,6)
35,000( P F ,15%,6)
5.48 In comparing alternatives I and J by the present
worth method, the equation that yields the present
worth of alternative I is:
( a) PW I 150,000 11,000( P A ,15%,3)
25,000( P F ,15%,3)
(b) PW I 150,000 11,000( P A ,15%,6)
25,000( P F ,15%,6)
(c) PW I 150,000 11,000( P A ,15%,6)
175,000( P F ,15%,3) 25,000( P F ,15%,6)
(d) PW I 150,000 11,000( P A ,15%,6)
125,000( P F ,15%,3) 25,000( P F ,15%,6)
Problems 5.49 and 5.50 are based on the following
information.
Machine X Machine Y
Initial cost, $ 80,000 95,000
Annual operating cost, $ per year 20,000 15,000
Salvage value, $ 10,000 30,000
Life, years 2 4
The interest rate is 10% per year.
5.49 The equation that will calculate the present worth
of machine X is:
(a) PW X 80,000 15,000( P A ,10%,4)
30,000( P F ,10%,4)
(b) PW X 80,000 20,000( P A ,10%,4)
80,000( P F ,10%,2) 10,000( P F ,10%,4)
( c) PW X 80,000 20,000( P A ,10%,2)
10,000( P F ,10%,2)
(d) PW X 80,000 20,000( P A ,10%,4)
70,000( P F ,10%,2) 10,000( P F ,10%,4)
5.50 In comparing the machines on a present worth
basis, the present worth of machine Y is closest to:
(a) $112,320
(b) $122,060
(c) $163,040
(d) $175,980

Case Study 149
5.51 The capitalized cost of $10,000 every 5 years
forever , starting now at an interest rate of 10% per
year, is closest to:
( a) $13,520
( b) $16,380
( c) $26,380
( d) $32,590
5.52 At an interest rate of 10% per year, the capitalized
cost of $10,000 in year 0, $5000 per year in years 1
through 5, and $1000 per year thereafter forever is
closest to:
(a) $29,652
(b) $35,163
(c) $38,954
(d) $43,221
CASE STUDY
COMPARING SOCIAL SECURITY BENEFITS
Background
When Sheryl graduated from Northeastern University in
2000 and went to work for BAE Systems, she did not pay
much attention to the monthly payroll deduction for social
security. It was a “necessary evil” that may be helpful in retirement
years. However, this was so far in the future that she
fully expected this government retirement benefit system to
be broke and gone by the time she could reap any benefits
from her years of contributions.
This year, Sheryl and Brad, another engineer at BAE,
got married. Recently, they both received notices from the
Social Security Administration of their potential retirement
amounts, were they to retire and start social security benefits
at preset ages. Since both of them hope to retire a few
years early, they decided to pay closer attention to the predicted
amount of retirement benefits and to do some analysis
on the numbers.
Information
They found that their projected benefits are substantially the
same, which makes sense since their salaries are very close to
each other. Although the numbers were slightly different in
their two mailings, the similar messages to Brad and Sheryl
can be summarized as follows:
If you stop working and start receiving benefits . . .
At age 62, your payment would
be about
At you full retirement age (67
years), your payment would be
about
At age 70, your payment would
be about
$1400 per month
$2000 per month
$2480 per month
These numbers represent a reduction of 30% for early retirement
(age 62) and an increase of 24% for delayed retirement
(age 70).
This couple also learned that it is possible for a spouse to
take spousal benefits at the time that one of them is at full
retirement age. In other words, if Sheryl starts her $2000 benefit
at age 67, Brad can receive a benefit equal to 50% of hers.
Then, when Brad reaches 70 years of age, he can discontinue
spousal benefits and start his own. In the meantime, his benefits
will have increased by 24%. Of course, this strategy
could be switched with Brad taking his benefits and Sheryl
receiving spousal benefits until age 70.
All these options led them to define four alternative plans.
A: Each takes early benefits at age 62 with a 30% reduction
to $1400 per month.
B: Each takes full benefits at full retirement age of 67
and receives $2000 per month.
C: Each delays benefits until age 70 with a 24% increase
to $2480 per month.
D: One person takes full benefits of $2000 per month at
age 67, and the other person receives spousal benefits
($1000 per month at age 67) and switches to delayed
benefits of $2480 at age 70.
They realize, of course, that the numbers will change over
time, based on their respective salaries and number of years
of contribution to the social security system by them and by
their employers.
Case Study Exercises
Brad and Sheryl are the same age. Brad determined that most
of their investments make an average of 6% per year. With
this as the interest rate, the analysis for the four alternatives is
possible. Sheryl and Brad plan to answer the following questions,
but don’t have time this week. Can you please help
them? (Do the analysis for one person at a time, not the couple,
and stop at the age of 85.)
1. How much in total (without the time value of money considered)
will each plan A through D pay through age 85?
2. What is the future worth at 6% per year of each plan at
age 85?
3. Plot the future worth values for all four plans on one
spreadsheet graph.
4. Economically, what is the best combination of plans for
Brad and Sheryl, assuming they both live to be 85 years
old?
5. Develop at least one additional question that you think
Sheryl and Brad may have. Answer the question.

CHAPTER 6
Annual Worth
Analysis
LEARNING OUTCOMES
Purpose: Utilize different annual worth techniques to evaluate and select alternatives.
SECTION TOPIC LEARNING OUTCOME
6.1 Advantages of AW • Demonstrate that the AW value is the same for
each life cycle.
6.2 CR and AW values • Calculate and interpret the capital recovery (CR)
and AW amounts.
6.3 AW analysis • Select the best alternative using an annual worth
analysis.
6.4 Perpetual life • Evaluate alternatives with very long lives using
AW analysis.
6.5 LCC analysis • Perform a life-cycle cost (LCC) analysis using AW
methods.

I
n this chapter, we add to our repertoire of alternative comparison tools. In Chapter
5 we learned the PW method. Here we learn the equivalent annual worth, or
AW, method. AW analysis is commonly considered the more desirable of the two
methods because the AW value is easy to calculate; the measure of worth—AW in monetary
units per year—is understood by most individuals; and its assumptions are essentially identical
to those of the PW method.
Annual worth is also known by other titles. Some are equivalent annual worth (EAW),
equivalent annual cost (EAC), annual equivalent (AE), and equivalent uniform annual cost
(EUAC). The alternative selected by the AW method will always be the same as that selected
by the PW method, and all other alternative evaluation methods, provided they are performed
correctly.
An additional application of AW analysis treated here is life-cycle cost (LCC) analysis. This
method considers all costs of a product, process, or system from concept to phaseout.
6.1 Advantages and Uses of Annual Worth Analysis
For many engineering economic studies, the AW method is the best to use, when compared to
PW, FW, and rate of return (Chapters 7 and 8). Since the AW value is the equivalent uniform annual
worth of all estimated receipts and disbursements during the life cycle of the project or alternative,
AW is easy to understand by any individual acquainted with annual amounts, for example,
dollars per year. The AW value, which has the same interpretation as A used thus far, is
the economic equivalent of the PW and FW values at the MARR for n years. All three can be
easily determined from each other by the relation
AW PW( A P,i,n ) FW( A F,i,n ) [6.1]
The n in the factors is the number of years for equal-service comparison. This is the LCM or the
stated study period of the PW or FW analysis.
When all cash flow estimates are converted to an AW value, this value applies for every year
of the life cycle and for each additional life cycle.
The annual worth method offers a prime computational and interpretation advantage because
the AW value needs to be calculated for only one life cycle. The AW value determined over
one life cycle is the AW for all future life cycles. Therefore, it is not necessary to use the
LCM of lives to satisfy the equal-service requirement.
Equal-service
requirement and LCM
As with the PW method, there are three fundamental assumptions of the AW method that should
be understood. When alternatives being compared have different lives, the AW method makes the
assumptions that
1. The services provided are needed for at least the LCM of the lives of the alternatives.
2. The selected alternative will be repeated for succeeding life cycles in exactly the same manner
as for the first life cycle.
3. All cash flows will have the same estimated values in every life cycle.
In practice, no assumption is precisely correct. If, in a particular evaluation, the first two assumptions
are not reasonable, a study period must be established for the analysis. Note that for assumption
1, the length of time may be the indefinite future (forever). In the third assumption, all
cash flows are expected to change exactly with the inflation (or deflation) rate. If this is not a
reasonable assumption, new cash flow estimates must be made for each life cycle, and again a
study period must be used. AW analysis for a stated study period is discussed in Section 6.3.
EXAMPLE 6.1
In Example 5.3, National Homebuilders, Inc. evaluated cut-and-finish equipment from vendor A
(6-year life) and vendor B (9-year life). The PW analysis used the LCM of 18 years. Consider
only the vendor A option now. The diagram in Figure 6–1 shows the cash flows for all three life
cycles (first cost $15,000; annual M&O costs $3500; salvage value $1000). Demonstrate
the equivalence at i 15% of PW over three life cycles and AW over one cycle. In Example
5.3, present worth for vendor A was calculated as PW $45,036.

152 Chapter 6 Annual Worth Analysis
Figure 6–1
PW and AW values
for three life cycles,
Example 6.1.
PW = $45,036
3 life cycles
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
$1000
0 1 2 3 4 5 6
Life cycle 1
i = 15%
$3500
$1000
$15,000
0 1 2 3 4 5 6
Life cycle 2
$3500
$1000
$15,000
0 1 2 3 4 5 6
Life cycle 3
$3500
$15,000
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
• • •
continues
AW = $7349
Solution
Calculate the equivalent uniform annual worth value for all cash flows in the first life cycle.
AW 15,000( AP ,15%,6) 1000( AF ,15%,6) 3500 $7349
When the same computation is performed on each succeeding life cycle, the AW value is
$7349. Now Equation [6.1] is applied to the PW value for 18 years.
AW 45,036( AP ,15%,18) $7349
The one-life-cycle AW value and the AW value based on 18 years are equal.
Not only is annual worth an excellent method for performing engineering economy studies,
but also it is applicable in any situation where PW (and FW and benefit/cost) analysis can be
utilized. The AW method is especially useful in certain types of studies: asset replacement and
retention time studies to minimize overall annual costs (both covered in Chapter 11), breakeven
studies and make-or-buy decisions (Chapter 13), and all studies dealing with production or manufacturing
costs where a cost /unit or profit /unit measure is the focus.

6.2 Calculation of Capital Recovery and AW Values 153
If income taxes are considered, a slightly different approach to the AW method is used by some
large corporations and financial institutions. It is termed economic value added, or EVA. This approach,
covered in Chapter 17, concentrates upon the wealth-increasing potential that an alternative offers a
corporation. The resulting EVA values are the equivalent of an AW analysis of after-tax cash flows.
6.2 Calculation of Capital Recovery and AW Values
An alternative should have the following cash flow estimates:
Initial investment P. This is the total first cost of all assets and services required to initiate
the alternative. When portions of these investments take place over several years, their present
worth is an equivalent initial investment. Use this amount as P .
Salvage value S. This is the terminal estimated value of assets at the end of their useful life.
The S is zero if no salvage is anticipated; S is negative when it will cost money to dispose of
the assets. For study periods shorter than the useful life, S is the estimated market value or
trade-in value at the end of the study period.
Annual amount A. This is the equivalent annual amount (costs only for cost alternatives;
costs and receipts for revenue alternatives). Often this is the annual operating cost (AOC) or
M&O cost, so the estimate is already an equivalent A value.
Salvage/market value
The annual worth (AW) value for an alternative is comprised of two components: capital
recovery for the initial investment P at a stated interest rate (usually the MARR) and the equivalent
annual amount A. The symbol CR is used for the capital recovery component. In equation form,
AW CR A [6.2]
Both CR and A represent costs. The total annual amount A is determined from uniform recurring
costs (and possibly receipts) and nonrecurring amounts. The PA and PF factors may be necessary
to first obtain a present worth amount; then the AP factor converts this amount to the A
value in Equation [6.2]. (If the alternative is a revenue project, there will be positive cash flow
estimates present in the calculation of the A value.)
The recovery of an amount of capital P committed to an asset, plus the time value of the
capital at a particular interest rate, is a fundamental principle of economic analysis .
Capital recovery (CR) is the equivalent annual amount that the asset, process, or system must
earn (new revenue) each year to just recover the initial investment plus a stated rate of return
over its expected life. Any expected salvage value is considered in the computation of CR.
Capital recovery
The AP factor is used to convert P to an equivalent annual cost. If there is some anticipated
positive salvage value S at the end of the asset’s useful life, its equivalent annual value is recovered
using the AF factor. This action reduces the equivalent annual cost of the asset. Accordingly,
CR is calculated as
CR P ( A P,i,n ) S ( AF,i,n ) [6.3]
EXAMPLE 6.2
Lockheed Martin is increasing its booster thrust power in order to win more satellite launch contracts
from European companies interested in opening up new global communications markets. A
piece of earth-based tracking equipment is expected to require an investment of $13 million, with
$8 million committed now and the remaining $5 million expended at the end of year 1 of the
project. Annual operating costs for the system are expected to start the first year and continue at
$0.9 million per year. The useful life of the tracker is 8 years with a salvage value of $0.5 million.
Calculate the CR and AW values for the system, if the corporate MARR is 12% per year.
Solution
Capital recovery: Determine P in year 0 of the two initial investment amounts, followed by the
use of Equation [6.3] to calculate the capital recovery. In $1 million units,
P 8 5( PF ,12%,1) $12.46

154 Chapter 6 Annual Worth Analysis
CR 12.46( AP ,12%,8) 0.5( AF ,12%,8)
12.46(0.20130) 0.5(0.08130)
$2.47
The correct interpretation of this result is very important to Lockheed Martin. It means that
each and every year for 8 years, the equivalent total net revenue from the tracker must be at
least $2,470,000 just to recover the initial present worth investment plus the required return of
12% per year. This does not include the AOC of $0.9 million each year.
Annual worth: To determine AW, the cash flows in Figure 6–2a must be converted to an
equivalent AW series over 8 years ( Figure 6–2b ). Since CR $2.47 million is an equivalent
annual cost, as indicated by the minus sign, total AW is determined.
AW 2.47 0.9 $3.37 million per year
This is the AW for all future life cycles of 8 years, provided the costs rise at the same rate as
inflation, and the same costs and services are expected to apply for each succeeding life cycle.
$0.5
0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8
$0.9
AW=?
$5.0
Figure 6–2
( a ) Cash flow diagram
for satellite tracker
costs, and ( b ) conversion
to an equivalent
AW (in $1 million),
Example 6.2.
$8.0
(a)
(b)
There is a second, equally correct way to determine CR. Either method results in the same
value. There is a relation between the AP and AF factors.
( AF,i,n ) ( AP,i,n ) i
Both factors are present in the CR Equation [6.3]. Substitute for the AF factor to obtain
CR P ( AP,i,n ) S [( AP,i,n ) i ]
[( P S )( AP,i,n ) S ( i )]
There is a basic logic to this formula. Subtracting S from the initial investment P before applying
the AP factor recognizes that the salvage value will be recovered. This reduces CR, the annual
cost of asset ownership. However, the fact that S is not recovered until year n of ownership is
compensated for by charging the annual interest S ( i ) against the CR. Although either CR relation
results in the same amount, it is better to consistently use the same method. The first method,
Equation [6.3], will be used in this text.
For solution by spreadsheet, use the PMT function to determine CR only in a single spreadsheet
cell. The general function PMT( i %, n,P,F ) is rewritten using the initial investment as
P and S for the salvage value. The format is
PMT( i %, n,P , S ) [6.4]
As an illustration, determine only the CR in Example 6.2. The equivalent initial investment
in year 0 is $12.46 million. The complete function for the CR amount (in $1 million units) is
PMT(12%,8,12.46,0.5). The answer of $2.47 (million) will be displayed in the spreadsheet
cell.
As we learned in Section 3.1, one spreadsheet function can be embedded in another function.
In the case of Example 6.2, the initial investment is distributed over a 2-year period. The one-cell
PMT function, with the PV function embedded (in bold), can be written as PMT(12%,8,
8PV(12%,1, 5) ,0.5) to display the same value CR $2.47.

6.3 Evaluating Alternatives by Annual Worth Analysis 155
6.3 Evaluating Alternatives by Annual
Worth Analysis
The annual worth method is typically the easiest to apply of the evaluation techniques when the
MARR is specified. The AW is calculated over the respective life of each alternative, and the selection
guidelines are the same as those used for the PW method. For mutually exclusive alternatives,
whether cost- or revenue-based, the guidelines are as follows:
One alternative: If AW 0, the requested MARR is met or exceeded and the alternative is
economically justified.
Two or more alternatives: Select the alternative with the AW that is numerically largest,
that is, less negative or more positive. This indicates a lower AW of cost for cost alternatives
or a larger AW of net cash flows for revenue alternatives.
Project evaluation
ME alternative
selection
If any of the three assumptions in Section 6.1 is not acceptable for an alternative, a study period
analysis must be used. Then the cash flow estimates over the study period are converted to AW
amounts. The following two examples illustrate the AW method for one project and two alternatives.
EXAMPLE 6.3
Heavenly Pizza, which is located in Toronto, fares very well with its competition in offering
fast delivery. Many students at the area universities and community colleges work part-time
delivering orders made via the web. The owner, Jerry, a software engineering graduate, plans
to purchase and install five portable, in-car systems to increase delivery speed and accuracy.
The systems provide a link between the web order-placement software and the On-Star system
for satellite-generated directions to any address in the area. The expected result is faster, friendlier
service to customers and larger income.
Each system costs $4600, has a 5-year useful life, and may be salvaged for an estimated
$300. Total operating cost for all systems is $1000 for the first year, increasing by $100 per
year thereafter. The MARR is 10%. Perform an annual worth evaluation for the owner that
answers the following questions. Perform the solution by hand and by spreadsheet.
(a) How much new annual net income is necessary to recover the investment at the MARR
of 10% per year?
(b) Jerry estimates increased net income of $6000 per year for all five systems. Is this project
financially viable at the MARR?
(c) Based on the answer in part ( b ), determine how much new net income Heavenly Pizza
must have to economically justify the project. Operating costs remain as estimated.
Solution by Hand
(a) The capital recovery amount calculated by Equation [6.3] answers the first question.
CR 5[4600( AP ,10%,5)] 5[300( AF ,10%,5)]
5[4600(0.26380)] 5[300(0.16380)]
$5822
The five systems must generate an equivalent annual new revenue of $5822 to recover the
initial investment plus a 10% per year return.
( b) Figure 6–3 presents the cash flows over 5 years. The annual operating cost series,
combined with the estimated $6000 annual income, forms an arithmetic gradient series
with a base amount of $5000 and G $100. The project is financially viable if
AW 0 at i 10% per year. Apply Equation [6.2], where A is the equivalent annual
net income series.
AW CR A 5822 5000 100( AG ,10%,5)
$1003
The system is not financially justified at the net income level of $6000 per year.

156 Chapter 6 Annual Worth Analysis
Figure 6–3
Cash flow diagram used to
compute AW, Example 6.3.
$6000
$1500
0
1 2 3 4
5
$1000
$1100
$1200
$1300
$1400
$23,000
(c) Let the required income equal R, and set the AW relation equal to zero to find the minimum
income to justify the system.
0 5822 ( R 1000) 100( AG ,10%,5)
R 5822 1000 100(1.8101)
$7003 per year
Solution by Spreadsheet
The spreadsheet in Figure 6–4 summarizes the estimates and answers the questions posed
for Heavenly Pizza with the same values that were determined in the hand solution.
Figure 6–4
Spreadsheet solution
of Example 6.3.
( a ) Capital recovery
in cell B16, (b) AW in
cell E17, and ( c ) Goal
Seek template and
outcome in cell B5.
PMT($B$2, 5, NPV($B$2,B10:B14)+B9)
PMT($B$2, 5, NPV($B$2,E10:E14)+E9)
(a) and (b)
(c) Required income to
justify system
(c)

6.4 AW of a Permanent Investment 157
Cell references are used in the spreadsheet functions to accommodate changes in estimated
values.
(a) The capital recovery CR $–5822 is displayed in column B using the PMT function
with an embedded NPV function, as shown in the cell tag.
( b) The annual worth AW $–1003 is displayed in column E using the PMT function
shown. The arithmetic gradient series of costs and estimated income of $6000 in columns
C and D, respectively, are added to obtain the net income necessary for the PMT
function.
( c) The minimum required income is determined in the lower part of Figure 6–4 . This is
easily accomplished by setting AW 0 (column E) in the Goal Seek tool and letting it
find the income per year of $7003 to balance the AW equation.
EXAMPLE 6.4
Luby’s Cafeterias is in the process of forming a separate business unit that provides meals to
facilities for the elderly, such as assisted care and long-term care centers. Since the meals are
prepared in one central location and distributed by trucks throughout the city, the equipment
that keeps food and drink cold and hot is very important. Michele is the general manager of this
unit, and she wishes to choose between two manufacturers of temperature retention units that
are mobile and easy to sterilize after each use. Use the cost estimates below to select the more
economic unit at a MARR of 8% per year.
Hamilton (H)
Infinity Care (IC)
Initial cost P, $ 15,000 20,000
Annual M&O, $year 6,000 9,000
Refurbishment cost, $ 0 2,000 every 4 years
Trade-in value S, % of P 20 40
Life, years 4 12
Solution
The best evaluation technique for these different-life alternatives is the annual worth method,
where AW is taken at 8% per year over the respective lives of 4 and 12 years.
AW H annual equivalent of P annual M&O annual equivalent of S
15,000( AP ,8%,4) 6000 0.2(15,000)( AF ,8%,4)
15,000(0.30192) 6000 3000(0.22192)
$9,863
AW IC annual equivalent of P annual M&O annual equivalent of refurbishment
annual equivalent of S
20,000( AP ,8%,12) 9000 2000[( PF ,8%,4) ( PF ,8%,8)]( AP ,8%,12)
0.4(20,000)( AF ,8%,12)
20,000(0.13270) 9000 2000[0.7350 0.5403](0.13270) 8000(0.05270)
$11,571
The Hamilton unit is considerably less costly on an annual equivalent basis.
If the projects are independent, the AW at the MARR is calculated. All projects with AW 0
are acceptable.
6.4 AW of a Permanent Investment
This section discusses the annual worth equivalent of the capitalized cost introduced in Section
5.5. Evaluation of public sector projects, such as flood control dams, irrigation canals,
bridges, or other large-scale projects, requires the comparison of alternatives that have such long
Independent project
selection

158 Chapter 6 Annual Worth Analysis
lives that they may be considered infinite in economic analysis terms. For this type of analysis,
the annual worth (and capital recovery amount) of the initial investment is the perpetual annual
interest on the initial investment, that is, A Pi (CC) i. This is Equation [5.2].
Cash flows recurring at regular or irregular intervals are handled exactly as in conventional
AW computations; convert them to equivalent uniform annual amounts A for one cycle. This
automatically annualizes them for each succeeding life cycle. Add all the A values to the CR
amount to find total AW, as in Equation [6.2].
EXAMPLE 6.5
The U.S. Bureau of Reclamation is considering three proposals for increasing the capacity of the
main drainage canal in an agricultural region of Nebraska. Proposal A requires dredging the
canal to remove sediment and weeds that have accumulated during previous years’ operation.
The capacity of the canal will have to be maintained in the future near its design peak flow because
of increased water demand. The Bureau is planning to purchase the dredging equipment
and accessories for $650,000. The equipment is expected to have a 10-year life with a $17,000
salvage value. The annual operating costs are estimated to total $50,000. To control weeds in the
canal itself and along the banks, environmentally safe herbicides will be sprayed during the irrigation
season. The yearly cost of the weed control program is expected to be $120,000.
Proposal B is to line the canal with concrete at an initial cost of $4 million. The lining is
assumed to be permanent, but minor maintenance will be required every year at a cost of
$5000. In addition, lining repairs will have to be made every 5 years at a cost of $30,000.
Proposal C is to construct a new pipeline along a different route. Estimates are an initial cost
of $6 million, annual maintenance of $3000 for right-of-way, and a life of 50 years.
Compare the alternatives on the basis of annual worth, using an interest rate of 5% per
year.
Solution
Since this is an investment for a permanent project, compute the AW for one cycle of all recurring
costs. For proposals A and C, the CR values are found using Equation [6.3], with n A 10
and n C 50, respectively. For proposal B, the CR is simply P ( i ).
Proposal A
CR of dredging equipment:
650,000( AP ,5%,10) 17,000( AF ,5%,10) $ 82,824
Annual cost of dredging 50,000
Annual cost of weed control 120,000
$252,824
Proposal B
CR of initial investment: 4,000,000(0.05) $200,000
Annual maintenance cost 5,000
Lining repair cost: 30,000( AF ,5%,5) 5,429
$210,429
Proposal C
CR of pipeline: 6,000,000( AP ,5%,50)
$328,680
Annual maintenance cost 3,000
$331,680
Proposal B, which is a permanent solution, is selected due to its lowest AW of costs.
Comment
Note the use of the AF factor for the lining repair cost in proposal B. The AF factor is used
instead of AP , because the lining repair cost begins in year 5, not year 0, and continues
indefinitely at 5-year intervals.
If the 50-year life of proposal C is considered infinite, CR P ( i ) $300,000, instead of
$328,680 for n 50. This is a small economic difference. How long lives of 40 or more
years are treated economically is a matter of “local” practice.

6.4 AW of a Permanent Investment 159
EXAMPLE 6.6
At the end of each year, all owners and employees at Bell County Utility Cooperative are given
a bonus check based on the net profit of the Coop for the previous year. Bart just received his
bonus in the amount of $8530. He plans to invest it in an annuity program that returns
7% per year. Bart’s long-term plans are to quit the Coop job some years in the future when he is
still young enough to start his own business. Part of his future living expenses will be paid from
the proceeds that this year’s bonus accumulates over his remaining years at the Coop.
(a) Use a spreadsheet to determine the amount of annual year-end withdrawal that he can anticipate
(starting 1 year after he quits) that will continue forever. He is thinking of working
15 or 20 more years.
(b) Determine the amount Bart must accumulate after 15 and 20 years to generate $3000 per
year forever.
Solution by Spreadsheet
(a) Figure 6–5 presents the cash flow diagram for n 15 years of accumulation at 7% per
year on the $8530 deposited now. The accumulated amount after n 15 years is indicated
as F after 15 ? and the withdrawal series starts at the end of year 16. The diagram for
n 20 would appear the same, except there is a 20-year accumulation period.
The spreadsheet in Figure 6–6 shows the functions and answers for n 15 years in
columns C and D. The FV function displays the total accumulated at 7% after 15 years
as $23,535. The perpetual withdrawal is determined by viewing this accumulated amount
as a P value and by applying the formula
A P ( i ) 23,535(0.07) $1647 per year
The spreadsheet function D9*B7 performs this same computation in cell reference format
in column D.
Answers for n 20 years are displayed in column E. At a consistent 7% per year return,
Bart’s perpetual income is estimated as $1647 after 15 years, or $2311 per year if he
waits for 20 years.
i =7%
0 1 2 3 4 13 14 15
A =?
16 17 18 19 20 21 22 …
∞
P = $8530
F after 15 = ?
Figure 6–5
Diagram for a perpetual
series starting after 15
years of accumulation,
Example 6.6.
Figure 6–6
Spreadsheet solution,
Example 6.6.

160 Chapter 6 Annual Worth Analysis
(b) To obtain a perpetual annual withdrawal of $3000, it is necessary to determine how much
must be accumulated 1 year before the first withdrawal of $3000. This is an application of
the relation A P (i) solved for P , or
P — A i ——— 3000
0.07 $42,857
This P value is independent of how long Bart works at the Coop, because he must accumulate
this amount to achieve his goal. Figure 6–6 , row 13, shows the function and result. Note that
the number of years n does not enter into the function 3000B13.
Comment
The NPER function can be used to determine how many years it will take for the current
amount of $8530 to accumulate the required $42,857 at 7% per year. The row 15 function indicates
that Bart will have to work at the Coop for just under 24 additional years.
6.5 Life-Cycle Cost Analysis
The PW and AW analysis techniques discussed thus far have concentrated on estimates for first
cost P , annual operating and maintenance costs (AOC or M&O), salvage value S , and predictable
periodic repair and upgrade costs, plus any revenue estimates that may favor one alternative over
another. There are usually a host of additional costs involved when the complete project life costs
are evaluated. A life-cycle cost analysis includes these additional estimates to the extent that they
can be reliably determined.
Life-cycle cost (LCC) analysis utilizes AW or PW methods to evaluate cost estimates for the entire
life cycle of one or more projects. Estimates will cover the entire life span from the early
conceptual stage, through the design and development stages, throughout the operating stage, and
even the phaseout and disposal stages. Both direct and indirect costs are included to the extent
possible, and differences in revenue and savings projections between alternatives are included.
Some typical LCC applications are life-span analysis for military and commercial aircraft,
new manufacturing plants, new automobile models, new and expanded product lines, and government
systems at federal and state levels. For example, the U.S. Department of Defense
requires that a government contractor include an LCC budget and analysis in the originating
proposal for most defense systems.
Most commonly the LCC analysis includes costs, and the AW method is used for the analysis,
especially if only one alternative is evaluated. If there are expected revenue or other benefit differences
between alternatives, a PW analysis is recommended. Public sector projects are usually
evaluated using a benefit/cost analysis (Chapter 9), rather than LCC analysis, because estimates
to the citizenry are difficult to make with much accuracy. The direct costs mentioned above include
material, human labor, equipment, supplies, and other costs directly related to a product,
process, or system. Some examples of indirect cost components are taxes, management, legal,
warranty, quality, human resources, insurance, software, purchasing, etc. Direct and indirect
costs are discussed further in Chapter 15.
LCC analysis is most effectively applied when a substantial percentage of the life span (postpurchase)
costs, relative to the initial investment, will be expended in direct and indirect operating,
maintenance, and similar costs once the system is operational. For example, the evaluation
of two equipment purchase alternatives with expected useful lives of 5 years and M&O costs of
5% to 10% of the initial investment does not require an LCC analysis. However, let’s assume that
Exxon-Mobil wants to evaluate the design, construction, operation, and support of a new type
and style of tanker that can transport oil over long distances of ocean. If the initial costs are in the
$100 millions with support and operating costs ranging from 25% to 35% of this amount over a
25-year life, the logic of an LCC analysis will offer a better understanding of the economic viability
of the project.
To understand how a LCC analysis works, first we must understand the phases and stages of
systems engineering or systems development. Many books and manuals are available on systems
development and analysis. Generally, the LCC estimates may be categorized into a simplified

6.5 Life-Cycle Cost Analysis 161
format for the major phases of acquisition, operation, and phaseout/disposal , and their respective
stages.
Acquisition phase: all activities prior to the delivery of products and services.
• Requirements definition stage—Includes determination of user/customer needs, assessing
them relative to the anticipated system, and preparation of the system requirements
documentation.
• Preliminary design stage—Includes feasibility study, conceptual, and early-stage plans;
final go–no go decision is probably made here.
• Detailed design stage—Includes detailed plans for resources—capital, human, facilities,
information systems, marketing, etc.; there is some acquisition of assets, if economically
justifiable.
Operation phase: all activities are functioning, products and services are available.
• Construction and implementation stage—Includes purchases, construction, and implementation
of system components; testing; preparation, etc.
• Usage stage—Uses the system to generate products and services; the largest portion of
the life cycle.
Phaseout and disposal phase: covers all activities to transition to a new system; removal/
recycling/disposal of old system.
EXAMPLE 6.7
In the 1860s, General Mills Inc. and Pillsbury Inc. both started in the flour business in the Twin
Cities of Minneapolis–St. Paul, Minnesota. In the decade of 2000 to 2010, General Mills purchased
Pillsbury for a combination cash and stock deal worth more than $10 billion and integrated
the product lines. Food engineers, food designers, and food safety experts made many
cost estimates as they determined the needs of consumers and the combined company’s ability
to technologically and safely produce and market new food products. At this point only cost
estimates have been addressed—no revenues or profits.
Assume that the major cost estimates below have been made based on a 6-month study
about two new products that could have a 10-year life span for the company. Use LCC analysis
at the industry MARR of 18% to determine the size of the commitment in AW terms.
(Time is indicated in product-years. Since all estimates are for costs, they are not preceded by
a minus sign.)
Consumer habits study (year 0)
Preliminary food product design (year 1)
Preliminary equipment/plant design (year 1)
Detail product designs and test marketing (years 1, 2)
Detail equipment/plant design (year 2)
Equipment acquisition (years 1 and 2)
Current equipment upgrades (year 2)
New equipment purchases (years 4 and 8)
Annual equipment operating cost (AOC) (years 3–10)
$0.5 million
0.9 million
0.5 million
1.5 million each year
1.0 million
$2.0 million each year
1.75 million
2.0 million (year 4)
10% per purchase
thereafter
200,000 (year 3)
4% per year thereafter
Marketing, year 2
$8.0 million
years 3–10 5.0 million (year 3)
and 0.2 million
per year thereafter
year 5 only
3.0 million extra
Human resources, 100 new employees for 2000 hours
per year (years 3–10)
Phaseout and disposal (years 9 and 10)
$20 per hour (year 3)
5% per year
$1.0 million each year

162 Chapter 6 Annual Worth Analysis
Solution
LCC analysis can get complicated rapidly due to the number of elements involved. Calculate
the PW by phase and stage, add all PW values, then find the AW over 10 years. Values are in
$1 million units.
Acquisition phase:
Requirements definition: consumer study
PW $0.5
Preliminary design: product and equipment
PW 1.4( PF ,18%,1) $1.187
Detailed design: product and test marketing, and equipment
PW 1.5( PA ,18%,2) 1.0( PF ,18%,2) $3.067
Operation phase:
Construction and implementation: equipment and AOC
PW 2.0( PA ,18%,2) 1.75( PF ,18%,2) 2.0( PF ,18%,4) 2.2( PF ,18%,8)
Use: marketing
1 (——
1.04
0.2
1.18 ) 8
————— ( PF ,18%,2) $6.512
0.14
PW 8.0( PF ,18%,2) [5.0( PA ,18%,8) 0.2( PG ,18%,8)]( PF ,18%,2)
3.0( PF ,18%,5)
$20.144
Use: human resources: (100 employees)(2000 hyr)($20h) $4.0 million in year 3
1 (——
1.05
1.18 ) 8
PW 4.0 ————— ( PF ,18%,2) $13.412
0.13
Phaseout phase:
PW 1.0( PA ,18%,2)( PF ,18%,8) $0.416
The sum of all PW of costs is PW $45.238 million. Finally, determine the AW over the
expected 10-year life span.
AW 45.238 million( AP ,18%,10) $10.066 million per year
This is the LCC estimate of the equivalent annual commitment to the two proposed
products.
Often the alternatives compared by LCC do not have the same level of output or amount of
usage. For example, if one alternative will produce 20 million units per year and a second alternative
will operate at 35 million per year, the AW values should be compared on a currency unit/
unit produced basis, such as dollar/unit or euro/hour operated.
Figure 6–7 presents an overview of how costs may be distributed over an entire life cycle. For
some systems, typically defense systems, operating and maintenance costs rise fast after acquisition
and remain high until phaseout occurs.
The total LCC for a system is established or locked in early in the life cycle. It is not unusual
to have 75% to 85% of the entire life span LCC committed during the preliminary and detail

6.5 Life-Cycle Cost Analysis 163
Costs
Requirements
and
design
Acquisition and
installation
Operating and
maintenance
Phaseout
and disposal
Start
Time
End of
life cycle
Figure 6–7
Typical distribution of life-cycle costs of the phases for one
life cycle.
D
%
B
%
B
Reduced
total LCC
Committed
Committed
Cumulative LCC
#1
Actual
Cumulative LCC
#2
#1
E
F
C
A
Time
A
Time
Acquisition
phase
Operation
phase
Acquisition
phase
Operation
phase
(a)
Figure 6–8
LCC envelopes for committed and actual costs: ( a ) design 1, ( b ) improved design 2.
(b)
design stages. As shown in Figure 6–8 a , the actual or observed LCC (bottom curve AB ) will trail
the committed LCC throughout the life span (unless some major design flaw increases the total
LCC of design #1 above point B ).
The potential for signifi cantly reducing total LCC occurs primarily during the early stages. A
more effective design and more efficient equipment can reposition the envelope to design #2 in
Figure 6–8 b . Now the committed LCC curve AEC is below AB at all points, as is the actual LCC
curve AFC . It is this lower envelope #2 we seek. The shaded area represents the reduction in
actual LCC.
Even though an effective LCC envelope may be established early in the acquisition phase,
it is not uncommon that unplanned cost-saving measures are introduced during the acquisition
phase and early operation phase. These apparent “savings” may actually increase the total
LCC, as shown by curve AFD . This style of ad hoc cost savings, often imposed by management
early in the design stage and/or construction stage, can substantially increase costs later,
especially in the after-sale portion of the use stage. For example, the use of inferior-strength
concrete and steel has been the cause of structural failures many times, thus increasing the
overall life span LCC.

164 Chapter 6 Annual Worth Analysis
CHAPTER SUMMARY
The annual worth method of comparing alternatives is often preferred to the present worth method,
because the AW comparison is performed for only one life cycle. This is a distinct advantage when
comparing different-life alternatives. AW for the first life cycle is the AW for the second, third, and
all succeeding life cycles, under certain assumptions. When a study period is specified, the AW
calculation is determined for that time period, regardless of the lives of the alternatives.
For infinite-life (perpetual) alternatives, the initial cost is annualized simply by multiplying P
by i. For finite-life alternatives, the AW through one life cycle is equal to the perpetual equivalent
annual worth.
Life-cycle cost analysis is appropriate for systems that have a large percentage of costs in
operating and maintenance. LCC analysis helps in the analysis of all costs from design to
operation to disposal phases.
PROBLEMS
Annual Worth Calculations
6.1 If you are asked to provide an annual worth (AW)
comparison of alternatives after a present worth
(PW) comparison has already been done, what factor
multiplied by the PW values provide the correct
AW values?
6.2 List three assumptions that are inherent in the
annual worth method of comparing alternatives.
6.3 In the annual worth method of comparing alternatives
that have different lives, why do you calculate
the AW of the alternatives over their respective
life cycles instead of over the least common
multiple of their lives?
6.4 James developed the two cash flow diagrams shown
at the bottom of this page. The cash flows for alternative
B represent two life cycles of A. Calculate
the annual worth value of each over the
respective life cycles to demonstrate that they are
the same. Use an interest rate of 10% per year.
6.5 An asset with a first cost of $20,000 has an annual
operating cost of $12,000 and a $4000 salvage value
after its 4-year life. If the project will be needed for
6 years, what would the market (salvage) value of
the 2-year-old asset have to be for the annual worth
to be the same as it is for one life cycle of the asset?
Use an interest rate of 10% per year.
6.6 A sports mortgage is an innovative way to finance
cash-strapped sports programs by allowing fans to
sign up to pay a “mortgage” for the right to buy
good seats at football games for several decades
with season tickets locked in at current prices. At
Notre Dame, the locked-in price period is 50 years.
If a fan pays a $130,000 “mortgage” fee now (i.e.,
in year 0) when season tickets are selling for $290
each, what is the equivalent annual cost of the
football tickets over the 50-year period at an interest
rate of 8% per year?
6.7 If a fan buys a sports mortgage to USC football
games by paying $130,000 in 10 equal payments
starting now , and then pays a fixed price of $290
per year for 50 years (starting 1 year from now)
for season tickets, what is the AW in years 1
through 50 of the season tickets at 8% per year
interest?
Alternative A
Alternative B
$1000
i = 10% per year
$1000
0 1 2 3
Year
0
1
2 3 4 5 6
Year
$25 $25 $25
$25 $25 $25
$25 $25 $25
$5000
$5000
$4000

Problems 165
6.8 Eight years ago, Ohio Valley Trucking purchased a
large-capacity dump truck for $115,000 to provide
short-haul earthmoving services. The company
sold it today for $45,000. Operating and maintenance
costs averaged $10,500 per year. A complete
overhaul at the end of year 4 cost an extra $3600.
Calculate the annual cost of the truck at 8% per
year interest.
6.9 A major repair on the suspension system of a
5-year-old car cost $2000 because the warranty
expired after 3 years of ownership. The cost of
periodic maintenance has been $800 every 2 years.
If the owner donates the car to charity after 8 years
of ownership, what is the equivalent annual cost of
the repair and maintenance in the 8-year period
of ownership? Use an interest rate of 8% per year,
and assume that the owner paid the $800 maintenance
cost immediately before donating the car in
year 8.
Capital Recovery
6.10 Ten years ago, Jacobson Recovery purchased a
wrecker for $285,000 to move disabled 18-wheelers.
He anticipated a salvage value of $50,000 after 10
years. During this time his average annual revenue
totaled $52,000. ( a ) Did he recover his investment
and a 12% per year return? ( b ) If the annual M&O
cost was $10,000 the first year and increased by a
constant $1000 per year, was the AW positive or
negative at 12% per year? Assume the $50,000 salvage
was realized.
6.11 Sylvia has received a $500,000 inheritance from
her favorite, recently deceased aunt in Hawaii.
Sylvia is planning to purchase a condo in Hawaii
in the same area where her aunt lived all her life
and to rent it to vacationers. She hopes to make 8%
per year on this purchase over an ownership period
of 20 years. The condo’s total first cost is $500,000,
and she conservatively expects to sell it for 90% of
the purchase price. No annual M&O costs are considered
in the analysis. ( a ) What is the capital recovery
amount? ( b ) If there is a real boom in rental
real estate 10 years in the future, what sales price
(as a percentage of original purchase price) is necessary
at that time (year 10) to realize the same
amount as the 8% return expected over the 20-year
ownership period?
6.12 Humana Hospital Corporation installed a new MRI
machine at a cost of $750,000 this year in its medical
professional clinic in Cedar Park. This state-ofthe-art
system is expected to be used for 5 years
and then sold for $75,000. Humana uses a return
requirement of 24% per year for all of its medical
diagnostic equipment. As a bioengineering student
currently serving a coop semester on the management
staff of Humana Corporation in Louisville,
Kentucky, you are asked to determine the minimum
revenue required each year to realize the expected
recovery and return. Also, you are asked to
draw two cash flow diagrams, one showing the
MRI purchase and sale cash flow and a second depicting
the required capital recovery each year.
Alternative Comparison
6.13 Polypropylene wall caps, used for covering exterior
vents for kitchen cooktops, bathroom fans,
dryers, and other building air exhausts, can be
made by two different methods. Method X will
have a first cost of $75,000, an operating cost of
$32,000 per year, and a $9000 salvage value after
4 years. Method Y will have a first cost of $140,000,
an operating cost of $24,000 per year, and a
$19,000 salvage value after its 4-year life. At an
interest rate of 10% per year, which method should
be used on the basis of an annual worth analysis?
6.14 Nissan’s all-electric car, the Leaf, has a base price
of $32,780 in the United States, but it is eligible
for a $7500 federal tax credit. A consulting engineering
company wants to evaluate the purchase
or lease of one of the vehicles for use by its employees
traveling to job sites in the local area. The
cost for leasing the vehicle will be $4200 per year
(payable at the end of each year) after an initialization
charge of $2500 paid now. If the company
purchases the vehicle, it will also purchase a home
charging station for $2200 that will be partially
offset by a 50% tax credit. If the company expects
to be able to sell the car and charging station for
40% of the base price of the car alone at the end of
3 years, should the company purchase or lease the
car? Use an interest rate of 10% per year and annual
worth analysis.
6.15 A new structural design software package is available
for analyzing and designing three-sided guyed
towers and three- and four-sided self-supporting
towers. A single-user license will cost $6000 per
year. A site license has a one-time cost of $22,000.
A structural engineering consulting company is
trying to decide between two alternatives: buy a
single-user license now and one each year for the
next 3 years (which will provide 4 years of service),
or buy a site license now. Determine which
strategy should be adopted at an interest rate of
10% per year for a 4-year planning period using
the annual worth method of evaluation.
6.16 The city council in a certain southwestern city is
considering whether to construct permanent restrooms
in 22 of its smaller parks (i.e., parks of less

166 Chapter 6 Annual Worth Analysis
than 12 acres) or pay for portable toilets on a yearround
basis. The cost of constructing the 22 permanent
restrooms will be $3.8 million. The 22
portable restrooms can be rented for $7500 each
for 1 year. The service life of a permanent restroom
is 20 years. Using an interest rate of 6% per
year and an annual worth analysis, determine if the
city should build the permanent restrooms or lease
the portable ones.
6.17 A remotely located air sampling station can be powered
by solar cells or by running an above ground
electric line to the site and using conventional
power. Solar cells will cost $16,600 to install and
will have a useful life of 5 years with no salvage
value. Annual costs for inspection, cleaning, etc.,
are expected to be $2400. A new power line will
cost $31,000 to install, with power costs expected
to be $1000 per year. Since the air sampling project
will end in 5 years, the salvage value of the line is
considered to be zero. At an interest rate of 10% per
year, ( a ) which alternative should be selected on the
basis of an annual worth analysis and ( b ) what must
be the first cost of the above ground line to make the
two alternatives equally attractive economically?
6.18 The cash flows for two small raw water treatment
systems are shown. Determine which should be
selected on the basis of an annual worth analysis at
10% per year interest.
MF UF
First cost, $ –33,000 –51,000
Annual cost, $ per year –8,000 –3,500
Salvage value, $ 4,000 11,000
Life, years 3 6
6.19 PGM Consulting is under contract to Montgomery
County for evaluating alternatives that use a
robotic, liquid-propelled “pig” to periodically inspect
the interior of buried potable water pipes for
leakage, corrosion, weld strength, movement over
time, and a variety of other parameters. Two
equivalent robot instruments are available. Robot
Joeboy will have a first cost of $85,000, annual
M&O costs of $30,000, and a $40,000 salvage
value after 3 years. Robot Watcheye will have a
first cost of $125,000, annual M&O costs of
$27,000, and a $33,000 salvage value after its
5-year life. Assume an interest rate of 8% per year.
(a)
(b)
Which robot is the better economic option?
Using the spreadsheet Goal Seek tool, determine
the first cost of the robot not selected in
( a ) so that it will be the economic selection.
6.20 TT Racing and Performance Motor Corporation
wishes to evaluate two alternative CNC machines
for NHRA engine building. Use the AW method at
10% per year to select the better alternative.
Machine R
Machine S
First cost, $ 250,000 370,500
Annual operating cost, 40,000 –50,000
$ per year
Salvage value, $ 20,000 30,000
Life, years 3 5
6.21 The maintenance and operation (M&O) cost of
front-end loaders working under harsh environmental
conditions tends to increase by a constant
$1200 per year for the first 5 years of operation.
For a loader that has a first cost of $39,000 and
first-year M&O cost of $17,000, compare the
equivalent annual worth of a loader kept for 4 years
with one kept for 5 years at an interest rate of
12% per year. The salvage value of a used loader is
$23,000 after 4 years and $18,000 after 5 years.
6.22 You work for Midstates Solar Power. A manager
asked you to determine which of the following
two machines will have the lower ( a ) capital recovery
and ( b ) equivalent annual total cost. Machine
Semi2 has a first cost of $80,000 and an
operating cost of $21,000 in year 1, increasing
by $500 per year through year 5, after which
time it will have a salvage value of $13,000. Machine
Auto1 has a first cost of $62,000 and an
operating cost of $21,000 in year 1, increasing
by 8% per year through year 5, after which time
it will have a scavenge value of $2000. Utilize an
interest rate of 10% per year to determine both
estimates.
Permanent Investments
6.23 The State of Chiapas, Mexico, decided to fund a
program for literacy. The first cost is $200,000
now, and an update budget of $100,000 every
7 years forever is requested. Determine the perpetual
equivalent annual cost at an interest rate of
10% per year.
6.24 Calculate the perpetual equivalent annual cost
(years 1 through infinity) of $5 million in year 0,
$2 million in year 10, and $100,000 in years 11
through infinity. Use an interest rate of 10% per
year.
6.25 A Pennsylvania coal mining operation has installed
an in-shaft monitoring system for oxygen
tank and gear readiness for emergencies. Based on
maintenance patterns for previous systems, there
are no maintenance costs for the first 2 years, they

Problems 167
increase for a time period, and then they level off.
Maintenance costs are expected to be $150,000 in
year 3, $175,000 in year 4, and amounts increasing
by $25,000 per year through year 6 and remain
constant thereafter for the expected 10-year
life of the system. If similar systems with similar
costs will replace the current one, determine the
perpetual equivalent annual maintenance cost at
i 10% per year.
6.26 Compare two alternatives for a security system
surrounding a power distribution substation using
annual worth analysis and an interest rate of 10%
per year.
Condi
Torro
First cost, $ 25,000 130,000
Annual cost, $ per year 9,000 2,500
Salvage value, $ 3,000 150,000
Life, years 3
6.27 A new bridge across the Allegheny River in
Pittsburgh is expected to be permanent and will
have an initial cost of $30 million. This bridge must
be resurfaced every 5 years at a cost of $1 million.
The annual inspection and operating costs are estimated
to be $50,000. Determine its equivalent annual
worth at an interest rate of 10% per year.
6.28 For the cash flows shown, use an annual worth
comparison and an interest rate of 10% per year.
( a) Determine the alternative that is economically
best.
( b) Determine the first cost required for each of
the two alternatives not selected in ( a ) so that
all alternatives are equally acceptable. Use a
spreadsheet to answer this question.
X Y Z
First cost, $ 90,000 400,000 650,000
Annual cost, 40,000 20,000 13,000
$ per year
Overhaul every — — 80,000
10 years, $
Salvage value, $ 7,000 25,000 200,000
Life, years 3 10
Life-Cycle Cost
6.29 Blanton Agriculture of Santa Monica, California,
offers different types and levels of irrigation
water conservation systems for use in areas where
groundwater depletion is a serious concern. A
large farming corporation in India, where depletion
is occurring at an alarming rate of 1.6 inches
(4 centimeters) per year due to exponential growth
in usage for crop irrigation, is considering the
purchase of one of the Blanton systems. There are
three options with two levels of automation for
the first two options. The estimated costs and associated
cash flow diagrams over a 10-year period
are summarized below and on the next page,
respectively, for each of the five alternatives.
Costs are categorized as design (Des), development
(Dev), and operation (Oper). For alternatives
A and B, there is an extra cost of $15,000 per
installation year to maintain the manual system in
place now. Level 2 development costs are distributed
equally over a 2-year period. Alternative C
is a retrofit of the current manual system with no
design or development costs, and there is no level
1 option. At an interest rate of 10% per year and a
10-year study period, determine which alternative
and level has the lowest LCC.
Alternative
Cost
Component
Level
1
Level
2
A Des $100,000 $200,000
Dev 175,000 350,000
Oper 60,000 55,000
Install time 1 year 2 years
B Des $ 50,000 $100,000
Dev 200,000 500,000
Oper 45,000 30,000
Install time 1 year 2 years
C Oper $100,000
6.30 The Pentagon asked a defense contractor to estimate
the life-cycle cost for a proposed light-duty
support vehicle. The list of items included the following
categories: R&D costs (R&D), nonrecurring
investment costs (NRI), recurring upgrade costs
(RU), scheduled and unscheduled maintenance
costs (Maint), equipment usage costs (Equip), and
phaseoutdisposal costs (PoD). Use the cost estimates
(shown in $1 million) for the 20-year life
cycle to calculate the annual LCC at an interest rate
of 7% per year.
Year R&D NRI RU Maint Equip PoD
0 5.5 1.1
1 3.5
2 2.5
3 0.5 5.2 1.3 0.6 1.5
4 10.5 3.1 1.4 3.6
5 10.5 4.2 1.6 5.3
6–10 6.5 2.7 7.8
11 on 2.2 3.5 8.5
18–20 2.7

168 Chapter 6 Annual Worth Analysis
Alternative A
Level 1
0 1 2 3 4 5 6 7 8 9 10
$100,000
$60,000
$175,000
15,000
Alternative A
Level 2
0 1 2 3 4 5 6 7 8 9 10
$55,000
$200,000
$175,000
15,000
0 1 2 3 4 5 6 7 8 9 10
Alternative B
Level 1
$50,000
$45,000
$200,000
15,000
Alternative B
Level 2
0 1 2 3 4 5 6 7 8 9 10
$30,000
$100,000
$250,000
15,000
Alternative C
Level 2
0 1 2 3 4 5 6 7 8 9 10
Cash flow diagrams for Problem 6.29
$100,000
6.31 A manufacturing software engineer at a major
aerospace corporation has been assigned the
management responsibility of a project to design,
build, test, and implement AREMSS, a
new- generation automated scheduling system
for routine and expedited maintenance. Reports
on the disposition of each service will also be
entered by field personnel, then filed and archived
by the system. The initial application will
be on existing Air Force in-flight refueling aircraft.
The system is expected to be widely used
over time for other aircraft maintenance scheduling.
Once it is fully implemented, enhancements
will have to be made, but the system is expected
to serve as a worldwide scheduler for up to
15,000 separate aircraft. The engineer, who must
make a presentation next week of the best estimates
of costs over a 20-year life period, has decided
to use the life-cycle cost approach of cost
estimations. Use the following information to
determine the current annual LCC at 6% per year
for the AREMSS scheduling system.

Additional Problems and FE Exam Review Questions 169
Cost in Year ($ millions)
Cost Category 1 2 3 4 5 6 on 10 18
Field study 0.5
Design of system 2.1 1.2 0.5
Software design 0.6 0.9
Hardware purchases 5.1
Beta testing 0.1 0.2
Users manual
0.1 0.1 0.2 0.2 0.06
development
System implementation 1.3 0.7
Field hardware 0.4 6.0 2.9
Training trainers 0.3 2.5 2.5 0.7
Software upgrades 0.6 3.0 3.7
6.32 The U.S. Army received two proposals for a turnkey
design-build project for barracks for infantry unit soldiers
in training. Proposal A involves an off-the-shelf
“bare-bones” design and standard grade construction
of walls, windows, doors, and other features. With
this option, heating and cooling costs will be greater,
maintenance costs will be higher, and replacement
will be sooner than for proposal B. The initial cost for
A will be $750,000. Heating and cooling costs will
average $72,000 per year, with maintenance costs
averaging $24,000 per year. Minor remodeling will
be required in years 5, 10, and 15 at a cost of $150,000
each time in order to render the units usable for
20 years. They will have no salvage value.
Proposal B will include tailored design and
construction costs of $1.1 million initially, with
estimated heating and cooling costs of $36,000 per
year and maintenance costs of $12,000 per year.
There will be no salvage value at the end of the
20-year life. Which proposal should be accepted
on the basis of an annual life-cycle cost analysis, if
the interest rate is 6% per year?
6.33 A medium-size municipality plans to develop a
software system to assist in project selection during
the next 10 years. A life-cycle cost approach
has been used to categorize costs into development,
programming, operating, and support costs
for each alternative. There are three alternatives
under consideration, identified as M, N, and O.
The costs are summarized below. Use an annual
life-cycle cost approach to identify the best alternative
at 8% per year.
Alternative Component Estimated Cost
M Development $250,000 now,
$150,000 years 1–4
Programming $45,000 now,
$35,000 years 1, 2
Operation $50,000 years 1–10
Support $30,000 years 1–5
N Development $10,000 now
Programming $45,000 year 0,
$30,000 years 1–3
Operation $80,000 years 1–10
Support $40,000 years 1–10
O Operation $175,000 years 1–10
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
6.34 All of the following are fundamental assumptions
for the annual worth method of analysis except :
( a) The alternatives will be needed for only one
life cycle.
( b) The services provided are needed for at least
the LCM of the lives of the alternatives.
( c) The selected alternative will be repeated for
the succeeding life cycles in exactly the same
manner as for the first life cycle.
( d) All cash flows will have the same estimated
values in every life cycle.
6.35 When comparing five alternatives that have different
lives by the AW method, you must:
( a) Find the AW of each over the life of the
longest-lived alternative.
( b) Find the AW of each over the life of the
shortest-lived alternative.
( c) Find the AW of each over the LCM of all of
the alternatives.
( d) Find the AW of each alternative over its life
without considering the life of the other
alternatives.
6.36 The annual worth of an alternative can be calculated
from the alternative’s:
( a) Present worth by multiplying by ( A/P , i , n )
(b) Future worth by multiplying by ( F/A , i , n )
(c) Either ( a ) or ( b )
(d) Neither ( a ) nor ( b )
6.37 The alternatives shown are to be compared on the
basis of annual worth. At an interest rate of 10%
per year, the values of n that you could use in the
( AP,i,n ) factors to make a correct comparison by
the annual worth method are:
First cost, $ 50,000 90,000
Annual cost, $ per year 10,000 4,000
Salvage value, $ 13,000 15,000
Life, years 3 6
(a) n 3 years for A and 3 years for B
(b) n 3 years for A and 6 years for B
(c) Either ( a ) or ( b )
(d) Neither ( a ) nor ( b )
A
B

170 Chapter 6 Annual Worth Analysis
6.38 The alternatives shown are to be compared on the
basis of a perpetual (i.e., forever) equivalent annual
worth. At an interest rate of 10% per year, the
equation that represents the perpetual AW of X1 is:
First cost, $ 50,000 90,000
Annual cost, $ per year 10,000 4,000
Salvage value, $ 13,000 15,000
Life, years 3 6
X1
(a) AW X1 50,000(0.10) 10,000
13,000 (0.10)
(b) AW X1 50,000(0.10) 10,000
13,000 ( A/F ,10%,3)
(c) AW X1 50,000(0.10) 10,000
37,000 ( P/F ,10%,3)(0.10)
13,000(0.10)
(d) AW X1 50,000( A/P ,10%,3) 10,000
13,000( A/F ,10%,3)
6.39 To get the AW of a cash flow of $10,000 that
occurs every 10 years forever, with the first one
occurring 10 years from now, you should:
(a) Multiply $10,000 by ( A/P,i ,10).
(b) Multiply $10,000 by ( A/F,i ,10).
(c) Multiply $10,000 by i .
(d) Multiply $10,000 by ( A/F,i,n ) and then
multiply by i .
Problems 6.40 through 6.43 refer to the following
estimates.
The alternatives are mutually exclusive and the MARR is
6% per year.
Y1
Vendor 1 Vendor 2 Vendor 3
First cost, $ 200,000 550,000 1,000,000
Annual cost, $ per year 50,000 20,000 10,000
Revenue, $ per year 120,000 120,000 110,000
Salvage value, $ 25,000 0 500,000
Life, years 10 15
6.40 The annual worth of vendor 2 cash flow estimates
is closest to:
(a) $63,370
(b) $43,370
(c) $43,370
(d) $63,370
6.41 Of the following three relations, the correct one or ones
to calculate the annual worth of vendor 1 cash flow estimates
is ( note : all dollar values are in thousands):
Relation 1: AW 1 200( AP ,6%,10) 70
25( AF ,6%,10)
Relation 2: AW 1 [200 50( PA ,6%,10)
120( PA ,6%,10)
25( PF ,6%,10)](AP ,6%,10)
Relation 3:
(a) 1 and 3
(b) Only 1
(c) 1 and 2
(d) Only 3
AW 1 200( FP ,6%,10) 25
(50 120)( AP ,6%,10)
6.42 The AW values for the alternatives are listed
below. The vendor or vendors that should be recommended
is:
AW 1 $44,723 AW 2 $43,370
AW 3 $40,000
(a) 1 and 2
(b) 3
(c) 2
(d) 1
6.43 The capital recovery amount for vendor 3 is:
(a) $40,000 per year
(b) $60,000 per year
(c) $43,370 per year
(d) $100,000 per year
6.44 If a revenue alternative has a negative AW value and
it was correctly calculated, it means the following:
(a) The equivalent annual worth of revenues
does not exceed that of the costs.
(b)
(c)
The estimates are wrong somewhere.
A minus or plus sign of a cash flow was entered
incorrectly into the PMT spreadsheet
function.
(d) The alternative should have a longer life so
revenues will exceed costs.
6.45 Estimates for one of two process upgrades are as
follows: first cost of $40,000, annual cost of $5000
per year, market value that decreases by $2000 per
year to the salvage value of $20,000 after the expected
life of 10 years. If a 4-year study period is
used for AW analysis at 15% per year, the correct
AW value is closest to:
(a) $15,000
(b) $11,900
(c) $7600
(d) $12,600
6.46 The perpetual annual worth of investing $50,000
now and $20,000 per year starting in year 16 and
continuing forever at 12% per year is closest to:
(a) $4200
(b) $8650
(c) $9655
(d) $10,655

Case Study 171
6.47 All the following statements about the capital recovery
amount for an alternative are false except:
( a) Annual revenue can be no more than this
amount, if the alternative is selected.
( b) A monetary estimate of new capital funds required
each year for the life of the alternative.
(c) An amount of revenue required to recover
the first cost plus a stated return over the life
of the alternative.
(d) Does not consider the salvage value, since it
is returned at the end of the alternative’s life.
CASE STUDY
THE CHANGING SCENE OF AN ANNUAL WORTH ANALYSIS
Background and Information
Harry, owner of an automobile battery distributorship in
Atlanta, Georgia, performed an economic analysis 3 years ago
when he decided to place surge protectors in-line for all his
major pieces of testing equipment. The estimates used and the
annual worth analysis at MARR 15% are summarized
below. Two different manufacturers’ protectors were compared.
PowrUp
Lloyd’s
Cost and installation, $ 26,000 36,000
Annual maintenance cost, 800 300
$ per year
Salvage value, $ 2,000 3,000
Equipment repair savings, $ 25,000 35,000
Useful life, years 6 10
The spreadsheet in Figure 6–9 is the one Harry used to make
the decision. Lloyd’s was the clear choice due to its substantially
larger AW value. The Lloyd’s protectors were installed.
During a quick review this last year (year 3 of operation),
it was obvious that the maintenance costs and repair savings
have not followed (and will not follow) the estimates made
3 years ago. In fact, the maintenance contract cost (which includes
quarterly inspection) is going from $300 to $1200 per
year next year and will then increase 10% per year for the next
10 years. Also, the repair savings for the last 3 years were
$35,000, $32,000, and $28,000, as best as Harry can determine.
He believes savings will decrease by $2000 per year
hereafter. Finally, these 3-year-old protectors are worth nothing
on the market now, so the salvage in 7 years is zero, not
$3000.
Case Study Exercises
1. Plot a graph of the newly estimated maintenance costs
and repair savings projections, assuming the protectors
last for 7 more years.
2. With these new estimates, what is the recalculated AW
for the Lloyd’s protectors? Use the old first cost and
maintenance cost estimates for the first 3 years. If these
estimates had been made 3 years ago, would Lloyd’s
still have been the economic choice?
3. How has the capital recovery amount changed for the
Lloyd’s protectors with these new estimates?
Figure 6–9
Annual worth analysis of surge protector alternatives, case study.

CHAPTER 7
Rate of Return
Analysis: One
Project
LEARNING OUTCOMES
Purpose: Understand the meaning of rate of return and perform an ROR evaluation of a single project.
SECTION TOPIC LEARNING OUTCOME
7.1 Definition • State and understand the meaning of rate of
return.
7.2 Calculate ROR • Use a PW or AW relation to determine the
ROR of a series of cash flows.
7.3 Cautions • State the difficulties of using the ROR method,
relative to the PW and AW methods.
7.4 Multiple RORs • Determine the maximum number of possible
ROR values and their values for a specific cash
flow series.
7.5 Calculate EROR • Determine the external rate of return using
the techniques of modified ROR and return on
invested capital.
7.6 Bonds • Calculate the nominal and effective rate of
return for a bond investment.

T
he most commonly quoted measure of economic worth for a project or alternative
is its rate of return (ROR). Whether it is an engineering project with cash
flow estimates or an investment in a stock or bond, the rate of return is a wellaccepted
way of determining if the project or investment is economically acceptable. Compared
to the PW or AW value, the ROR is a generically different type of measure of worth,
as is discussed in this chapter. Correct procedures to calculate a rate of return using a PW or
AW relation are explained here, as are some cautions necessary when the ROR technique is
applied to a single project’s cash flows.
The ROR is known by other names such as the internal rate of return (IROR), which is the
technically correct term, and return on investment (ROI). We will discuss the computation of
ROI in the latter part of this chapter.
In some cases, more than one ROR value may satisfy the PW or AW equation. This chapter
describes how to recognize this possibility and an approach to find the multiple values .
Alternatively, one reliable ROR value can be obtained by using additional information established
separately from the project cash flows. Two of the techniques are covered: the modified
ROR technique and the ROIC (return on invested capital) technique.
Only one alternative is considered here; Chapter 8 applies these same principles to multiple
alternatives. Finally, the rate of return for a bond investment is discussed.
7.1 Interpretation of a Rate of Return Value
From the perspective of someone who has borrowed money, the interest rate is applied to the
unpaid balance so that the total loan amount and interest are paid in full exactly with the last loan
payment. From the perspective of a lender of money, there is an unrecovered balance at each time
period. The interest rate is the return on this unrecovered balance so that the total amount lent and
the interest are recovered exactly with the last receipt. Rate of return describes both of these
perspectives.
Rate of return (ROR) is the rate paid on the unpaid balance of borrowed money , or the rate
earned on the unrecovered balance of an investment , so that the final payment or receipt
brings the balance to exactly zero with interest considered.
Rate of return
The rate of return is expressed as a percent per period, for example, i 10% per year. It is
stated as a positive percentage; the fact that interest paid on a loan is actually a negative rate of
return from the borrower’s perspective is not considered. The numerical value of i can range from
100% to infinity, that is, 100% i . In terms of an investment, a return of i 100%
means the entire amount is lost.
The definition above does not state that the rate of return is on the initial amount of the investment;
rather it is on the unrecovered balance , which changes each time period. Example 7.1
illustrates this difference.
EXAMPLE 7.1
To get started in a new telecommuting position with AB Hammond Engineers, Jane took out
a $1000 loan at i 10% per year for 4 years to buy home office equipment. From the lender’s
perspective, the investment in this young engineer is expected to produce an equivalent net
cash flow of $315.47 for each of 4 years.
A $1000( A P ,10%,4) $315.47
This represents a 10% per year rate of return on the unrecovered balance. Compute the amount
of the unrecovered investment for each of the 4 years using ( a ) the rate of return on the unrecovered
balance (the correct basis) and ( b ) the return on the initial $1000 investment. ( c ) Explain
why all of the initial $1000 amount is not recovered by the final payment in part ( b ).
Solution
(a) Table 7–1 shows the unrecovered balance at the end of each year in column 6 using the
10% rate on the unrecovered balance at the beginning of the year. After 4 years the total
$1000 is recovered, and the balance in column 6 is exactly zero.

174 Chapter 7 Rate of Return Analysis: One Project
TABLE 7–1 Unrecovered Balances Using a Rate of Return of 10% on the Unrecovered
Balance
(1) (2) (3) 0.10 (2) (4) (5) (4) (3) (6) (2) (5)
Year
Beginning
Unrecovered
Balance
Interest on
Unrecovered
Balance
Cash
Flow
Recovered
Amount
Ending
Unrecovered
Balance
0 — — $1000.00 — $1000.00
1 $1000.00 $100.00 315.47 $215.47 784.53
2 784.53 78.45 315.47 237.02 547.51
3 547.51 54.75 315.47 260.72 286.79
4 286.79 28.68 315.47 286.79 0
$261.88 $1000.00
TABLE 7–2 Unrecovered Balances Using a 10% Return on the Initial Amount
(1) (2) (3) 0.10 (2) (4) (5) (4) (3) (6) (2) (5)
Year
Beginning
Unrecovered
Balance
Interest on
Initial Amount
Cash
Flow
Recovered
Amount
Ending
Unrecovered
Balance
0 — — $1000.00 — $1000.00
1 $1000.00 $100 315.47 $215.47 784.53
2 784.53 100 315.47 215.47 569.06
3 569.06 100 315.47 215.47 353.59
4 353.59 100 315.47 215.47 138.12
$400 $861.88
(b) Table 7–2 shows the unrecovered balance if the 10% return is always figured on the initial
$1000. Column 6 in year 4 shows a remaining unrecovered amount of $138.12,
because only $861.88 is recovered in the 4 years (column 5).
(c) As shown in column 3, a total of $400 in interest must be earned if the 10% return each year is
based on the initial amount of $1000. However, only $261.88 in interest must be earned if a
10% return on the unrecovered balance is used. There is more of the annual cash flow available
to reduce the remaining loan when the rate is applied to the unrecovered balance as in part ( a )
and Table 7–1 . Figure 7–1 illustrates the correct interpretation of rate of return in Table 7–1 .
Loan balance, $
1000.00
784.53
547.51
Loan balance
of $1000
$100.00
$215.47
Interest
Loan balance
reduction
$78.45
$237.02
$54.75
$260.72
286.79
$28.68
$286.79
0 1 2 3 4 Year
Loan balance
of $0
Figure 7–1
Plot of unrecovered balances and 10% per year rate of return on a $1000
amount, Table 7–1 .

7.2 Rate of Return Calculation Using a PW or AW Relation 175
Each year the $315.47 receipt represents 10% interest on the unrecovered balance in column 2
plus the recovered amount in column 5.
Because rate of return is the interest rate on the unrecovered balance, the computations in
Table 7–1 for part (a) present a correct interpretation of a 10% rate of return. Clearly, an interest
rate applied only to the principal represents a higher rate than is stated. In practice, a socalled
add-on interest rate is frequently based on principal only, as in part ( b ). This is sometimes
referred to as the installment fi nancing problem.
Installment fi nancing can be discovered in many forms in everyday finances. One popular
example is a “no-interest program” offered by retail stores on the sale of major appliances, audio
and video equipment, furniture, and other consumer items. Many variations are possible, but in
most cases, if the purchase is not paid for in full by the time the promotion is over, usually
6 months to 1 year later, fi nance charges are assessed from the original date of purchase. Further,
the program’s fine print may stipulate that the purchaser use a credit card issued by the retail
company, which often has a higher interest rate than that of a regular credit card, for example,
24% per year compared to 15% per year. In all these types of programs, the one common theme
is more interest paid over time by the consumer. Usually, the correct definition of i as interest on
the unpaid balance does not apply directly; i has often been manipulated to the financial disadvantage
of the purchaser. This was demonstrated by Example 4.4 using the Credit Card Case in
Chapter 4.
7.2 Rate of Return Calculation Using a PW
or AW Relation
The ROR value is determined in a generically different way compared to the PW or AW value for
a series of cash flows. For a moment, consider only the present worth relation for a cash flow
series. Using the MARR, which is established independent of any particular project’s cash flows,
a mathematical relation determines the PW value in actual monetary units, say, dollars or euros.
For the ROR values calculated in this and later sections, only the cash flows themselves are used
to determine an interest rate that balances the present worth relation. Therefore, ROR may be
considered a relative measure, while PW and AW are absolute measures. Since the resulting interest
rate depends only on the cash flows themselves, the correct term is internal rate of return
(IROR) ; however, the term ROR is used interchangeably. Another definition of rate of return is
based on our previous interpretations of PW and AW.
The rate of return is the interest rate that makes the present worth or annual worth of a cash
flow series exactly equal to 0.
Rate of return
To determine the rate of return, develop the ROR equation using either a PW or AW relation, set
it equal to 0, and solve for the interest rate. Alternatively, the present worth of cash outflows
(costs and disbursements) PW O may be equated to the present worth of cash inflows (revenues
and savings) PW I . That is, solve for i using either of the relations
or
0 PW [7.1]
PW O PW I
The annual worth approach utilizes the AW values in the same fashion to solve for i .
or
0 AW [7.2]
AW O AW I
The i value that makes these equations numerically correct is called i *. It is the root of the ROR
relation. To determine if the investment project’s cash flow series is viable, compare i * with the
established MARR.

176 Chapter 7 Rate of Return Analysis: One Project
Figure 7–2
Cash flow for which
a value of i is to be
determined.
$1500
$500
0 1 2 3 4 5
i * = ?
$1000
The guideline is as follows:
Project evaluation
If i * MARR, accept the project as economically viable.
If i * MARR, the project is not economically viable.
The purpose of engineering economy calculations is equivalence in PW or AW terms for a
stated i 0%. In rate of return calculations, the objective is to fi nd the interest rate i* at which
the cash flows are equivalent. The calculations are the reverse of those made in previous chapters,
where the interest rate was known. For example, if you deposit $1000 now and are promised payments
of $500 three years from now and $1500 five years from now, the rate of return relation
using PW factors and Equation [7.1] is
1000 500( P F , i *,3) 1500( P F , i *,5) [7.3]
The value of i * that makes the equality correct is to be determined (see Figure 7–2 ). If the $1000
is moved to the right side of Equation [7.3], we have the form 0 PW.
0 1000 500( P F , i *,3) 1500( P F , i *,5)
The equation is solved for i * 16.9% by hand using trial and error or using a spreadsheet function.
The rate of return will always be greater than zero if the total amount of cash inflow is greater
than the total amount of outflow, when the time value of money is considered. Using i * 16.9%,
a graph similar to Figure 7–1 can be constructed. It will show that the unrecovered balances each
year, starting with $1000 in year 1, are exactly recovered by the $500 and $1500 receipts in
years 3 and 5.
It should be evident that rate of return relations are merely a rearrangement of a present worth
equation. That is, if the above interest rate is known to be 16.9%, and it is used to find the present
worth of $500 three years from now and $1500 five years from now, the PW relation is
PW 500( P F ,16.9%,3) 1500( P F ,16.9%,5) $1000
This illustrates that rate of return and present worth equations are set up in exactly the same fashion.
The only differences are what is given and what is sought.
There are several ways to determine i * once the PW relation is established: solution via trial
and error by hand, using a programmable calculator, and solution by spreadsheet function. The
spreadsheet is faster; the first helps in understanding how ROR computations work. We summarize
two methods here and in Example 7.2 . Refer to Appendix D for the discussion about solution
by calculator.
i* Using Trial and Error The general procedure of using a PW-based equation is as follows:
1. Draw a cash flow diagram.
2. Set up the rate of return equation in the form of Equation [7.1] .
3. Select values of i by trial and error until the equation is balanced.
When the trial-and-error method is applied to determine i *, it is advantageous in step 3 to get
fairly close to the correct answer on the first trial. If the cash flows are combined in such a manner
that the income and disbursements can be represented by a single factor such as P F or P A , it is

7.2 Rate of Return Calculation Using a PW or AW Relation 177
possible to look up the interest rate (in the tables) corresponding to the value of that factor for n
years. The problem, then, is to combine the cash flows into the format of only one of the factors.
This may be done through the following procedure:
1. Convert all disbursements into either single amounts ( P or F ) or uniform amounts ( A ) by
neglecting the time value of money. For example, if it is desired to convert an A to an F
value, simply multiply the A by the number of years n . The scheme selected for movement
of cash flows should be the one that minimizes the error caused by neglecting the time value
of money. That is, if most of the cash flow is an A and a small amount is an F , convert the F
to an A rather than the other way around.
2. Convert all receipts to either single or uniform values.
3. Having combined the disbursements and receipts so that a P F , P A , or A F format applies,
use the interest tables to find the approximate interest rate at which the P F , P A , or A F
value is satisfied. The rate obtained is a good estimate for the first trial.
It is important to recognize that this first-trial rate is only an estimate of the actual rate of return,
because the time value of money is neglected. The procedure is illustrated in Example 7.2 .
i* by Spreadsheet The fastest way to determine an i * value when there is a series of equal
cash flows (A series) is to apply the RATE function. This is a powerful one-cell function, where it
is acceptable to have a separate P value in year 0 and a separate F value in year n . The format is
RATE (n,A,P,F) [7.4]
When cash flows vary from year to year (period to period), the best way to find i * is to enter
the net cash flows into contiguous cells (including any $0 amounts) and apply the IRR function
in any cell. The format is
IRR(first_cell:last_cell,guess) [7.5]
where “guess” is the i value at which the function starts searching for i *.
The PW-based procedure for sensitivity analysis and a graphical estimation of the i * value is
as follows:
1. Draw the cash flow diagram.
2. Set up the ROR relation in the form of Equation [7.1], PW 0.
3. Enter the cash flows onto the spreadsheet in contiguous cells.
4. Develop the IRR function to display i *.
5. Use the NPV function to develop a PW graph (PW versus i values). This graphically shows
the i * value at which PW 0.
EXAMPLE 7.2
Applications of green, lean manufacturing techniques coupled with value stream mapping can
make large financial differences over future years while placing greater emphasis on environmental
factors. Engineers with Monarch Paints have recommended to management an investment
of $200,000 now in novel methods that will reduce the amount of wastewater, packaging
materials, and other solid waste in their consumer paint manufacturing facility. Estimated savings
are $15,000 per year for each of the next 10 years and an additional savings of $300,000
at the end of 10 years in facility and equipment upgrade costs. Determine the rate of return
using hand and spreadsheet solutions.
Solution by Hand
Use the trial-and-error procedure based on a PW equation.
1. Figure 7–3 shows the cash flow diagram.
2. Use Equation [7.1] format for the ROR equation.
0 200,000 15,000( P A , i *,10) 300,000( P F , i *,10) [7.6]

178 Chapter 7 Rate of Return Analysis: One Project
i * = ?
$300,000
$15,000
0 1 2 3 4 5 6 7 8 9 10
$200,000
Figure 7–3
Cash flow diagram, Example 7.2 .
3. Use the estimation procedure to determine i for the first trial. All income will be regarded
as a single F in year 10 so that the P F factor can be used. The P F factor is selected because
most of the cash flow ($300,000) already fits this factor and errors created by
neglecting the time value of the remaining money will be minimized. Only for the first
estimate of i , define P $200,000, n 10, and F 10(15,000) 300,000 $450,000.
Now we can state that
200,000 450,000( P F , i ,10)
( P F , i ,10) 0.444
The roughly estimated i is between 8% and 9%. Use 9% as the first trial because this approximate
rate for the P F factor will be lower than the true value when the time value of money is
considered.
0 200,000 15,000( P A ,9%,10) 300,000( P F ,9%,10)
0 $22,986
The result is positive, indicating that the return is more than 9%. Try i 11%.
0 200,000 15,000( P A ,11%,10) 300,000( P F ,11%,10)
0 $6002
Since the interest rate of 11% is too high, linearly interpolate between 9% and 11%.
22,986 0
i * 9.00 ————————
22,986 (6002) (2.0)
9.00 1.58 10.58%
Solution by Spreadsheet
The fastest way to find i * is to use the RATE function ( Equation [7.4]). The entry
RATE(10,15000,200000,300000) displays i * 10.55% per year. It is equally correct to
use the IRR function. Figure 7–4 , column B, shows the cash flows and IRR(B2:B12)
function to obtain i *.
For a complete spreadsheet analysis, use the procedure outlined above.
1. Figure 7–3 shows cash flows.
2. Equation [7.6] is the ROR relation.
3. Figure 7–4 shows the net cash flows in column B.
4. The IRR function in cell B14 displays i * 10.55%.
5. To graphically observe i * 10.55%, column D displays the PW graph for different
i values. The NPV function is used repeatedly to calculate PW for the xy scatter
chart.

7.3 Special Considerations When Using the ROR Method 179
NPV(C4,$B$3:$B$12) + $B$2
IRR(B2:B12)
Figure 7–4
Spreadsheet to determine i * and develop a PW graph, Example 7.2 .
Just as i * can be found using a PW equation, it may equivalently be determined using an AW
relation. This method is preferred when uniform annual cash flows are involved. Solution by
hand is the same as the procedure for a PW-based relation, except Equation [7.2] is used. In the
case of Example 7.2 , i* 10.55% is determined using the AW-based relation.
0 200,000( AP,i *,10) 15,000 300,000( AF,i *,10)
The procedure for solution by spreadsheet is exactly the same as outlined above using the IRR
function. Internally, IRR calculates the NPV function at different i values until NPV 0. (There is no
equivalent way to utilize the PMT function, since it requires a fixed value of i to calculate an A value.)
7.3 Special Considerations When Using
the ROR Method
The rate of return method is commonly used in engineering and business settings to evaluate
one project, as discussed in this chapter, and to select one alternative from two or more, as
explained in the next chapter. As mentioned earlier, an ROR analysis is performed using a different
basis than PW and AW analyses. The cash flows themselves determine the (internal) rate
of return. As a result, there are some assumptions and special considerations with ROR analysis
that must be made when calculating i * and in interpreting its real-world meaning. A summary
is provided below.
• Multiple i * values. Depending upon the sequence of net cash inflows and outflows, there may
be more than one real-number root to the ROR equation, resulting in more than one i* value.
This possibility is discussed in Section 7.4.
• Reinvestment at i*. Both the PW and AW methods assume that any net positive investment
(i.e., net positive cash flows once the time value of money is considered) is reinvested at the
MARR. However, the ROR method assumes reinvestment at the i * rate. When i * is not close
to the MARR (e.g., if i * is substantially larger than MARR), this is an unrealistic assumption.
In such cases, the i * value is not a good basis for decision making. This situation is discussed
in Section 7.5.
• Different procedure for multiple alternative evaluations. To correctly use the ROR method
to choose from two or more mutually exclusive alternatives requires an incremental analysis
procedure that is significantly more involved than PW and AW analysis. Chapter 8 explains
this procedure.
If possible, from an engineering economic study perspective, the AW or PW method at a
stated MARR should be used in lieu of the ROR method . However, there is a strong appeal

180 Chapter 7 Rate of Return Analysis: One Project
for the ROR method because rate of return values are very commonly quoted. And it is easy to
compare a proposed project’s return with that of in-place projects.
When it is important to know the exact value of i*, a good approach is to determine PW or
AW at the MARR, then determine the specific i* for the selected alternative.
As an illustration, if a project is evaluated at MARR 15% and has PW 0, there is no need to
calculate i *, because i * 15%. However, if PW is positive, but close to 0, calculate the exact i *
and report it along with the conclusion that the project is financially justified.
7.4 Multiple Rate of Return Values
In Section 7.2 a unique rate of return i * was determined. In the cash flow series presented thus
far, the algebraic signs on the net cash fl ows changed only once, usually from minus in year 0 to
plus at some time during the series. This is called a conventional ( or simple) cash fl ow series.
However, for some series the net cash flows switch between positive and negative from one year
to another, so there is more than one sign change. Such a series is called nonconventional (nonsimple).
As shown in the examples of Table 7–3 , each series of positive or negative signs may be
one or more in length. Relatively large net cash flow (NCF) changes in amount and sign can
occur in projects that require significant spending at the end of the expected life. Nuclear plants,
open-pit mines, petroleum well sites, refineries, and the like often require environmental restoration,
waste disposal, and other expensive phaseout costs. The cash flow diagram will appear
similar to Figure 7–5 a . Plants and systems that have anticipated major refurbishment costs or
upgrade investments in future years may have considerable swings in cash flow and sign changes
over the years, as shown by the pattern in Figure 7–5 b .
TABLE 7–3 Examples of Conventional and Nonconventional Net Cash Flow for
a 6-year Project
Type of Series
Sign on Net Cash Flow by Year
0 1 2 3 4 5 6
Number of
Sign Changes
Conventional 1
Conventional 1
Conventional 1
Nonconventional 2
Nonconventional 2
Nonconventional 3
Positive
NCF
Positive
NCF
Positive
NCF
0 1 2 n 1 n
Year
0 1 2 n 1 n
Year
Initial
investment
i * = ?
Phaseout
costs
Initial
investment
Midlife
investments
(a)
(b)
Figure 7–5
Typical cash flow diagrams for projects with ( a ) large restoration or remediation costs, and ( b ) upgrade or refurbishment costs.
i * = ?

7.4 Multiple Rate of Return Values 181
When there is more than one sign change in the net cash flows, it is possible that there will be
multiple i * values in the 100% to plus infinity range. There are two tests to perform in sequence
on the nonconventional series to determine if there is one unique value or possibly multiple i *
values that are real numbers.
Test 1: (Descartes’) rule of signs states that the total number of real-number roots is always less
than or equal to the number of sign changes in the series.
This rule is derived from the fact that the relation set up by Equation [7.1] or [7.2] to find i *
is an n th-order polynomial. (It is possible that imaginary values or infinity may also satisfy the
equation.)
Test #2: Cumulative cash flow sign test, also known as Norstrom’s criterion, states that only
one sign change in a series of cumulative cash flows which starts negatively indicates that there
is one positive root to the polynomial relation.
Zero values in the series are neglected when applying Norstrom’s criterion. This is a more discriminating
test that determines if there is one, real-number, positive i * value. There may be
negative roots that satisfy the ROR relation, but these are not useful i * values. To perform the
test, determine the series
S t cumulative cash flows through period t
Observe the sign of S 0 and count the sign changes in the series S 0 , S 1 , . . . , S n . Only if S 0 0 and
signs change one time in the series is there a single, real-number, positive i *.
With the results of these two tests, the ROR relation is solved for either the unique i * or the
multiple i * values, using trial and error by hand, using a programmable calculator, or by spreadsheet
using an IRR function that incorporates the “guess” option. Development of the PW graph
is recommended, especially when using a spreadsheet. Examples 7.3 and 7.4 illustrate the tests
and solution for i *.
EXAMPLE 7.3
Sept-Îles Aluminum Company operates a bauxite mine to supply its aluminum smelter located
about 2 km from the current open pit. A new branch for the pit is proposed that will
supply an additional 10% of the bauxite currently available over the next 10-year period. The
lease for the land will cost $400,000 immediately. The contract calls for the restoration of the
land and development as part of a state park and wildlife area at the end of the 10 years. This
is expected to cost $300,000. The increased production capacity is estimated to net an additional
$75,000 per year for the company. Perform an ROR analysis that will provide the following
information:
(a) Type of cash flow series and possible number of ROR values
(b) PW graph showing all i* values
(c) Actual i* values determined using the ROR relation and spreadsheet function
(d) Conclusions that can be drawn about the correct rate of return from this analysis
Solution
(a) The net cash flows will appear like those in Figure 7–5a with an initial investment of
$400,000, annual net cash flow (NCF) of $75,000 for years 1 through 10, and a phaseout
cost of $300,000 in year 10. Figure 7–6 details the NCF series (column B) and cumulative
NCF (column C) for use in the two tests for unique and multiple i* values. The
series is nonconventional based on the sign changes throughout the series.
Test #1: There are two sign changes in the NCF series, which indicates a possible
maximum of two roots to the polynomial equation or i* values for the ROR
equation.
Test #2: There is one sign change in the cumulative NCF series, which indicates a unique
positive root or one positive i* value.

182 Chapter 7 Rate of Return Analysis: One Project
PV(D3,$B$4:$B$13)$B$3
IRR(B3:B13,guess)
Figure 7–6
Spreadsheet determination of multiple i* values and PW graph, Example 7.3.
(b) Columns D and E of the spreadsheet in Figure 7–6 use i values ranging from 20% to
20% per year to plot the PW vs i curve via the NPV function. There are two times that
the parabolic-shaped curve crosses the PW 0 line; these are approximately i 1 * 18%
and i 2 * 5%.
(c) The ROR equation based on PW computations is
0 400,000 75,000(PA,i*%,10) 300,000(PF,i*%,10) [7.7]
i* values by hand If hand solution is chosen, the same procedure used in Example 7.2
can be applied here. However, the technique to estimate the initial i value will not work
as well in this case since the majority of the cash flows do not fit either the PF or the
FP factor. In fact, using the PF factor, the initial i value is indicated to be 1.25%.
Trial-and-error solution of Equation [7.7] with various i values will approximate the
correct answer of about 4.5% per year. This complies with the test results of one positive
i* value.
i* by spreadsheet function Use the IRR(B3:B13,guess) function to determine the i*
value for the NCF series in column B, Figure 7–6. Entering different values in the
optional “guess” field will force the function to find multiple i* values, if they exist.
As shown in row 17, two are found.
i 1 * 18.70%
i 2 * 4.53%
This result does not conflict with test results, as there is one positive value, but a negative
value also balances the ROR equation.
(d) The positive i* 4.53% is accepted as the correct internal rate of return (IROR) for the
project. The negative value is not useful in economic conclusions about the project.
EXAMPLE 7.4
The engineering design and testing group for Honda Motor Corp. does contract-based work for
automobile manufacturers throughout the world. During the last 3 years, the net cash flows for
contract payments have varied widely, as shown below, primarily due to a large manufacturer’s
inability to pay its contract fee.
Year 0 1 2 3
Cash Flow ($1000) 2000 500 8100 6800

7.4 Multiple Rate of Return Values 183
(a) Determine the maximum number of i* values that may satisfy the ROR equation.
(b) Write the PW equation and approximate the i* value(s) by plotting PW vs i.
(c) What do the i* values mean?
Solution
(a) Table 7–4 shows the annual cash flows and cumulative cash flows. Since there are two
sign changes in the cash flow sequence, the rule of signs indicates a maximum of two i*
values. The cumulative cash flow sequence starts with a positive number S 0 2000,
indicating that test #2 is inconclusive. As many as two i* values can be found.
TABLE 7–4 Cash Flow and Cumulative Cash Flow Sequences, Example 7.4
Year
Cash Flow
($1000)
Sequence
Number
Cumulative Cash Flow
($1000)
0 2000 S 0 2000
1 500 S 1 1500
2 8100 S 2 6600
3 6800 S 3 200
(b) The PW relation is
PW 2000 500(PF,i,1) 8100(PF,i,2) 6800(PF,i,3)
The PW values are shown below and plotted in Figure 7–7 for several i values. The characteristic
parabolic shape for a second-degree polynomial is obtained, with PW crossing
the i axis at approximately i 1 * 8% and i 2 * 41%.
i% 5 10 20 30 40 50
PW ($1000) 51.44 39.55 106.13 82.01 11.83 81.85
100
75
50
Figure 7–7
Present worth of cash
flows at several interest
rates, Example 7.4.
PW ( $1000)
25
0
25
i%
10 20 30 40 50
50
75
100
125
Figure 7–8 presents the spreadsheet PW graph with the PW curve crossing the x axis at
PW 0 two times. Also, the solution for two positive i* values using the IRR function
with different guess values is displayed. The values are
i 1 * 7.47% i 2 * 41.35%
(c) Since both i* values are positive, they are not of much value, because neither can be
considered the true ROR of the cash flow series. This result indicates that additional

184 Chapter 7 Rate of Return Analysis: One Project
IRR function format
IRR($B4:$B7,guess)
Figure 7–8
Spreadsheet solution, Example 7.4.
information is needed to calculate a more useful project ROR, namely, some information
about the anticipated return on funds invested external to the project and the cost of capital
to borrow money to continue the project. This problem is a good example of when an
approach discussed in the next section should be taken.
If the guess option is not used in the IRR function, the starting point is 10% per year. The function
will find the one ROR closest to 10% that satisfies the PW relation. Entering various guess
values will allow IRR to find multiple i * values in the range of 100% to , if they exist. Often,
the results are unbelievable or unacceptable values that are rejected. Some helpful guidelines can
be developed. Assume there are two i * values for a particular cash flow series.
If the Results Are
Both i * 0
Both i * 0
One i * 0; one i * 0
What to Do
Discard both values.
Discard both values.
Use i * 0 as ROR.
If both i * values are discarded, proceed to the approach discussed in the next section to determine
one rate of return value for the project. However, remember the prime recommendation.
Always determine the PW or AW at the MARR first for a reliable measure of economic justification.
If the PW or AW is greater than zero and the ROR is needed, then find the actual i* of the
project cash flows.
This recommendation is not to dissuade you from using the ROR method. Rather it is a recommendation
that the use of the ROR technique be reserved for times when the actual i * value is
essential to the decision-making process.
7.5 Techniques to Remove Multiple Rates of Return
The techniques developed here are used under the following conditions:
• The PW or AW value at the MARR is determined and could be used to make the decision, but information
on the ROR is deemed necessary to finalize the economic decision, and
• The two tests of cash flow sign changes (Descartes’ and Norstrom’s) indicate multiple roots
( i * values) are possible, and

7.5 Techniques to Remove Multiple Rates of Return 185
• More than one positive i * value or all negative i * values are obtained when the PW graph and
IRR function are developed, and
• A single, reliable rate of return value is required by management or engineers to make a clear
economic decision.
We will present a couple of ways to remove multiple i * values. The selected approach depends
upon what estimates are the most reliable for the project being evaluated. An important fact to
remember is the following.
The result of follow-up analysis to obtain a single ROR value when multiple, nonuseful i* values
are present does not determine the internal rate of return (IROR) for nonconventional net
cash flow series. The resulting rate is a function of the additional information provided to make
the selected technique work, and the accuracy is further dependent upon the reliability of this
information.
We will refer to the resulting value as the external rate of return (EROR) as a reminder that
it is different from the IROR obtained in all previous sections. First, it is necessary to identify
the perspective about the annual net cash flows of a project. Take the following view: You are
the project manager and the project generates cash flows each year. Some years produce positive
NCF, and you want to invest the excess money at a good rate of return. We will call this
the investment rate i i . This can also be called the reinvestment rate . Other years, the net cash flow
will be negative and you must borrow funds from some source to continue. The interest rate you
pay should be as small as possible; we will call this the borrowing rate i b , also referred to as the
finance rate . Each year, you must consider the time value of money, which must utilize either the
investment rate or the borrowing rate, depending upon the sign on the NCF of the preceding year.
With this perspective, it is now possible to outline two approaches that rectify the multiple
i * situation. The resulting ROR value will not be the same for each method, because slightly
different additional information is necessary and the cash flows are treated in slightly different
fashions from the time value of money viewpoint.
Modified ROR (MIRR) Approach This is the easier approach to apply, and it has a spreadsheet
function that can find the single EROR value quickly. However, the investment and borrowing
rates must be reliably estimated, since the results may be quite sensitive to them. The
symbol i will identify the result.
Return on Invested Capital (ROIC) Approach Though more mathematically rigorous,
this technique provides a more reliable estimate of the EROR and it requires only the investment
rate i i . The symbol i is used to indentify the result.
Before covering the techniques, it would be good to review the material in Section 7.1, including
Example 7.1 . Though the i or i value determined here is not the ROR defined earlier in the
chapter, the concepts used to make the ending cash flow balance equal to zero are used.
Modified ROR Approach
The technique requires that two rates external to the project net cash flows be estimated.
• Investment rate i i is the rate at which extra funds are invested in some source external to the
project. This applies to all positive annual NCF. It is reasonable that the MARR is used for this
rate.
• Borrowing rate i b is the rate at which funds are borrowed from an external source to provide
funds to the project. This applies to all negative annual NCF. The weighted average cost of
capital (WACC) can be used for this rate.
It is possible to make all rates the same, that is, i i i b MARR WACC. However, this is not
a good idea as it implies that the company is willing to borrow funds and invest in projects at the
same rate. This implies no profit margin over time, so the company can’t survive for long using
this strategy. Commonly MARR WACC, so usually i i i b . (See Section 1.9 for a quick review
of MARR and WACC and Chapter 10 for a more detailed discussion of WACC.)

186 Chapter 7 Rate of Return Analysis: One Project
PW 0 at i b
Find FW n of all
NCF > 0 at investment rate i i
NCF 1
NCF 2
NCF 4
NCF n1
0 1 2 3 4 n 1 n
Year
NCF 3 NCF n
NCF 0
Find PW 0 of all
FW
NCF < 0 at borrowing rate i n at i i
b
Figure 7–9
Typical cash flow diagram to determine modified rate of return i.
Figure 7–9 is a reference diagram that has multiple i * values, since the net cash flows change
sign multiple times. The modified ROR method uses the following procedure to determine a
single external rate of return i and to evaluate the economic viability of the project.
1. Determine the PW value in year 0 of all negative NCF at the borrowing rate i b (lightly
shaded area and resulting PW 0 value in Figure 7–9 ).
2. Determine the FW value in year n of all positive NCF at the investment rate i i (darker shaded
area and resulting FW n value in Figure 7–9 ).
3. Calculate the modified rate of return i at which the PW and FW values are equivalent over
the n years using the following relation, where i is to be determined.
FW n PW 0 ( F P , i %, n ) [7.8]
If using a spreadsheet rather than hand computation, the MIRR function displays i directly with
the format
MIRR(first_cell:last_cell, i b , i i ) [7.9]
4. The guideline for economic decision making compares the EROR or i to MARR.
If i MARR, the project is economically justified.
If i MARR, the project is not economically justified.
As in other situations, on the rare occasion that i MARR, there is indifference to the project's
economic acceptability; however, acceptance is the usual decision.
EXAMPLE 7.5
The cash flows experienced by Honda Motors in Example 7.4 are repeated below. There are
two positive i * values that satisfy the PW relation, 7.47% and 41.35% per year. Use the modified
ROR method to determine the EROR value. Studies indicate that Honda has a WACC of
8.5% per year and that projects with an estimated return of less than 9% per year are routinely
rejected. Due to the nature of this contract business, any excess funds generated are expected
to earn at a rate of 12% per year.
Year 0 1 2 3
Net Cash Flow ($1000) 2000 500 8100 6800

7.5 Techniques to Remove Multiple Rates of Return 187
Solution by Spreadsheet
Using the information in the problem statement, the rate estimates are as follows:
MARR:
Investment rate, i i :
Borrowing rate, i b :
9% per year
12% per year
8.5% per year
The fast way to find i is with the MIRR function. Figure 7–10 shows the result of i ' 9.39%
per year. Since 9.39% MARR, the project is economically justified.
MIRR(B2:B5,E2,E3)
Figure 7–10
Spreadsheet application of MIRR function, Example 7.5 .
It is vital that the interpretation be correct. The 9.39% is not the internal rate of return
(IROR); it is the external ROR (EROR) based on the two external rates for investing and borrowing
money.
Solution by Hand
Figure 7–9 can serve as a reference as the procedure to find i ' manually is applied.
Step 1. Find PW 0 of all negative NCF at i b 8.5%.
PW 0 500( P F ,8.5%,1) 8100( P F ,8.5%,2)
$7342
Step 2. Find FW 3 of all positive NCF at i i 12%.
FW 3 2000( F P ,12%,3) 6800
$9610
Step 3. Find the rate i ' at which the PW and FW are equivalent.
PW 0 ( F P , i ,3) FW 3 0
7342(1 i ) 3 9610 0
i ' ( 9610 ———
7342 ) 13 1
0.939 (9.39%)
Step 4. Since i ' MARR of 9%, the project is economically justified using this EROR
approach.
Return on Invested Capital Approach
The definition of ROIC should be understood before we discuss the approach.
Return on invested capital (ROIC) is a rate-of-return measure of how effectively a project utilizes
the funds invested in it, that is, funds that remain internal to the project. For a corporation, ROIC
is a measure of how effectively it utilizes the funds invested in its operations, including facilities,
equipment, people, systems, processes, and all other assets used to conduct business.

188 Chapter 7 Rate of Return Analysis: One Project
The technique requires that the investment rate i i be estimated for excess funds generated in any
year that they are not needed by the project. The ROIC rate, which has the symbol i , is determined
using an approach called the net-investment procedure . It involves developing a series of future
worth ( F ) relations moving forward 1 year at a time. In those years that the net balance of the
project cash flows is positive (extra funds generated by the project), the funds are invested at the i i
rate. Usually, i i is set equal to the MARR. When the net balance is negative, the ROIC rate is used,
since the project keeps all of its funds internal to itself. The ROIC method uses the following procedure
to determine a single external rate of return i and to evaluate the economic viability of the
project. Remember that the perspective is that you are the project manager and when the project
generates extra cash flows, they are invested external to the project at the investment rate i i .
1. Develop a series of future worth relations by setting up the following relation for each year t
( t 1, 2, . . . , n years).
F t F t 1 (1 k ) NCF t [7.10]
where F t future worth in year t based on previous year and time value of money
NCF t net cash flow in year t
k { i i if F t1 0 (extra funds available)
i if F t1 0 (project uses all available funds)
2. Set the future worth relation for the last year n equal to 0, that is, F n 0, and solve for i to
balance the equation. The i value is the ROIC for the specified investment rate i i .
The F t series and solution for i in the F n 0 relation can become involved mathematically.
Fortunately, the Goal Seek spreadsheet tool can assist in the determination of i because
there is only one unknown in the F n relation and the target value is zero. (Both the hand
and spreadsheet solutions are demonstrated in Example 7.6 .)
3. The guideline for economic decision making is the same as above, namely,
If ROIC MARR, the project is economically justified.
If ROIC MARR, the project is not economically justified.
It is important to remember that the ROIC is an external rate of return dependent upon
the investment rate choice. It is not the same as the internal rate of return discussed at the beginning
of this chapter, nor is it the multiple rates, nor is it the MIRR rate found by the previous
method. This is a separate technique to find a single rate for the project.
EXAMPLE 7.6
Once again, we will use the cash flows experienced by Honda Motors in Example 7.4 (repeated
below). Use the ROIC method to determine the EROR value. The MARR is 9% per year, and
any excess funds generated by the project can earn at a rate of 12% per year.
Year 0 1 2 3
Net Cash Flow ($1000) 2000 500 8100 6800
Solution by Hand
The hand solution is presented first to provide the logic of the ROIC method. Use MARR
9% and i i 12% per year in the procedure to determine i , which is the ROIC. Figure 7–11
details the cash flows and tracks the progress as each F t is developed. Equation [7.10] is applied
to develop each F t .
Step 1. Year 0: F 0 $2000
Since F 0 0, externally invest in year 1 at i i 12%.
Year 1: F 1 2000(1.12) 500 $1740
Since F 1 0, use i i 12% for year 2.

7.5 Techniques to Remove Multiple Rates of Return 189
$6800 $6800 $6800
F 0 $2000 F 1 $1740
F 3 = 0
at i = ?
0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3
$500
Year
F 2 $6151
$8100
$8100
(a) (b) (c) (d)
Figure 7–11
Application of ROIC method at i i 12% per year: ( a ) original cash flow; equivalent form in ( b ) year 1, ( c ) year 2, and ( d ) year 3.
Year 2: F 2 1740(1.12) 8100 $6151
Now F 2 0, use i for year 3, according to Equation [7.10].
Year 3: F 3 6151(1 i ) 6800
This is the last year. See Figure 7–11 for equivalent net cash flow diagrams.
Go to step 2.
Step 2. Solve for i ROIC from F 3 0.
6151(1 i ) 6800 0
i 68006151 1
0.1055 (10.55%)
Step 3. Since ROIC > MARR 9%, the project is economically justified.
Solution by Spreadsheet
Figure 7–12 provides a spreadsheet solution. The future worth values F 1 through F 3 are determined
by the conditional IF statements in rows 3 through 5. The functions are shown in column
D. In each year, Equation [7.10] is applied. If there are surplus funds generated by the project,
F t –1 0 and the investment rate i i (in cell E7) is used to find F t . For example, because the F 1
value (in cell C3) of $1740 0, the time value of money for the next year is calculated at the
investment rate of 12% per year, as shown in the hand solution above for year 2.
Figure 7–12
Spreadsheet application of ROIC method using Goal Seek, Example 7.6 .

190 Chapter 7 Rate of Return Analysis: One Project
The Goal Seek template sets the F 3 value to zero by changing the ROIC value (cell E8). The
result is i ROIC 10.55% per year. As before, since 10.55% 9%, the MARR, the project
is economically justified.
Comment
Note that the rate by the ROIC method (10.55%) is different than the MIRR rate (9.39%). Also
these are both different than the multiple rates determined earlier (7.47% and 41.35%). This
shows how dependent the different methods are upon the additional information provided
when multiple i * rates are indicated by the two sign tests.
Now that we have learned two techniques to remove multiple i * values, there are some connections
between the multiple i * values, the external rate estimates, and the resulting external
rates ( i and i ) obtained by the two methods.
Modified ROR technique When both the borrowing rate i b and the investment rate i i are
exactly equal to any one of the multiple i * values, the rate i ' found by the MIRR function, or by
hand solution, will equal the i * value. That is, all four parameters have the same value.
If any i * i b i i , then i ' i *
ROIC technique Similarly, if the investment rate i i is exactly equal to any one of the multiple
i * values, the rate found by the Goal Seek tool, or by hand when the equation F n 0 is solved,
will be i i * value.
Finally, it is very important to remember the following fact.
None of the details of the modified ROR (MIRR) technique or the return on invested capital
(ROIC) technique are necessary if the PW or AW method of project evaluation is applied at a
specific MARR. When the MARR is established, this is, in effect, fixing the i* value. Therefore,
a definitive economic decision can be made directly from the PW or AW value.
7.6 Rate of Return of a Bond Investment
A time-tested method of raising capital funds is through the issuance of an IOU, which is financing
through debt, not equity (see Chapter 1). One very common form of IOU is a bond—a longterm
note issued by a corporation or a government entity (the borrower) to finance major projects.
The borrower receives money now in return for a promise to pay the face value V of the
bond on a stated maturity date. Bonds are usually issued in face value amounts of $1000, $5000,
or $10,000. Bond dividend I, also called bond interest , is paid periodically between the time the
money is borrowed and the time the face value is repaid. The bond dividend is paid c times per
year. Expected payment periods are usually semiannually or quarterly. The amount of interest is
determined using the stated dividend or interest rate, called the bond coupon rate b.
I
(face value) (bond coupon rate)
————————————————
number of payment periods per year
I Vb ——
c [7.11]
There are many types or classifications of bonds. Four general classifications are summarized
in Table 7–5 according to their issuing entity, fundamental characteristics, and example names or
purposes. For example, Treasury securities are issued in different monetary amounts ($1000 and
up) with varying periods of time to the maturity date (Bills up to 1 year; Notes for 2 to 10 years).
In the United States, Treasury securities are considered a very safe bond purchase because they are
backed with the “full faith and credit of the U.S. government.” The safe investment rate indicated
in Figure 16 as the lowest level for establishing a MARR is the coupon rate on a U.S. Treasury
security. Funds obtained through corporate bond issues are used for new product development,
facilities upgrade, expansion into international markets, and similar business ventures.

7.6 Rate of Return of a Bond Investment 191
TABLE 7–5
Classification and Characteristics of Bonds
Classification Issued by Characteristics Examples
Treasury securities Federal government Backed by faith and credit of the federal
government
Municipal Local governments Federal tax-exempt
Issued against taxes received
Mortgage Corporation Backed by specified assets or mortgage
Low ratelow risk on first mortgage
Foreclosure, if not repaid
Debenture Corporation Not backed by collateral, but by
reputation of corporation
Bond rate may “float”
Higher interest rates and higher risks
EXAMPLE 7.7
General Electric just released $10 million worth of $10,000 ten-year bonds. Each bond pays dividends
semiannually at a rate of 6% per year. (a) Determine the amount a purchaser will receive each
6 months and after 10 years. (b) Suppose a bond is purchased at a time when it is discounted by 2%
to $9800. What are the dividend amounts and the final payment amount at the maturity date?
Solution
(a) Use Equation [7.11] for the dividend amount.
10,000 (0.06)
I —————— $300 per 6 months
2
The face value of $10,000 is repaid after 10 years.
(b) Purchasing the bond at a discount from face value does not change the dividend or final repayment
amounts. Therefore, $300 per 6 months and $10,000 after 10 years remain the amounts.
Bills ( 1 year)
Notes (2–10 years)
Bonds (10–30 years)
General obligation
Revenue
Zero coupon
Put
First mortgage
Second mortgage
Equipment trust
Convertible
Subordinated
Junk or high yield
The cash flow series for a bond investment is conventional and has one unique i *, which is
best determined by solving a PW-based rate of return equation in the form of Equation [7.1] , that
is, 0 PW.
EXAMPLE 7.8
Allied Materials needs $3 million in debt capital for expanded composites manufacturing.
It is offering small-denomination bonds at a discount price of $800 for a 4% $1000 bond
that matures in 20 years with a dividend payable semiannually. What nominal and effective
interest rates per year, compounded semiannually, will Allied Materials pay an investor?
Solution
The income that a purchaser will receive from the bond purchase is the bond dividend I $20 every
6 months plus the face value in 20 years. The PW-based equation for calculating the rate of return is
0 800 20(PA,i*,40) 1000(PF,i*,40)
Solve by the IRR function or by hand to obtain i* 2.8435% semiannually. The nominal interest
rate per year is computed by multiplying i* by 2.
Nominal i (2.8435)(2) 5.6870% per year, compounded semiannually
Using Equation [4.5], the effective annual rate is
i a (1.028435) 2 1 5.7678%

192 Chapter 7 Rate of Return Analysis: One Project
EXAMPLE 7.9
Gerry is a project engineer. He took a financial risk and bought a bond from a corporation that
had defaulted on its interest payments. He paid $4240 for an 8% $10,000 bond with dividends
payable quarterly. The bond paid no interest for the first 3 years after Gerry bought it. If interest
was paid for the next 7 years and then Gerry was able to resell the bond for $11,000, what rate
of return did he make on the investment? Assume the bond is scheduled to mature 18 years
after he bought it. Perform hand and spreadsheet analysis.
Solution by Hand
The bond interest received in years 4 through 10 was
I ——————
(10,000)(0.08) $200 per quarter
4
The effective rate of return per quarter can be determined by solving the PW equation developed
on a per quarter basis.
0 4240 200(PA,i* per quarter,28)(PF,i* per quarter,12)
11,000(PF,i* per quarter,40)
The equation is correct for i* 4.1% per quarter, which is a nominal 16.4% per year, compounded
quarterly.
Solution by Spreadsheet
Once all the cash flows are entered into contiguous cells, the function IRR(B2:B42) is used
in Figure 7–13, row 43, to display the answer of a nominal rate of return of 4.10% per quarter.
(Note that many of the row entries have been hidden to conserve space.) This is the same as the
nominal annual rate of
i* 4.10%(4) 16.4% per year, compounded quarterly
Gerry did well on his bond investment.
Figure 7–13
Spreadsheet solution for
a bond investment,
Example 7.9.
If a bond investment is being considered and a required rate of return is stated, the same
PW-based relation used to find i * can be used to determine the maximum amount to pay for
the bond now to ensure that the rate is realized. The stated rate is the MARR, and the PW
computations are performed exactly as they were in Chapter 5. As an illustration, in the last
example , if 12% per year, compounded quarterly, is the target MARR, the PW relation is
used to find the maximum that Gerry should pay now; P is determined to be $6004. The
quarterly MARR is 12%4 3%.
0 P 200( P A ,3%,28)( P F ,3%,12) 11,000( P F ,3%,40)
P $6004

Problems 193
CHAPTER SUMMARY
The rate of return of a cash flow series is determined by setting a PW-based or AW-based relation
equal to zero and solving for the value of i *. The ROR is a term used and understood by almost
everybody. Most people, however, can have considerable difficulty in calculating a rate of return
correctly for anything other than a conventional cash flow series. For some types of series, more
than one ROR possibility exists. The maximum number of i * values is equal to the number of
changes in the sign of the net cash flow series (Descartes’ rule of signs). Also, a single positive
rate can be found if the cumulative net cash flow series starts negatively and has only one sign
change (Norstrom’s criterion).
When multiple i * values are indicated, either of the two techniques covered in this chapter can
be applied to find a single, reliable rate for the nonconventional net cash flow series. In the case
of the ROIC technique, additional information is necessary about the investment rate that excess
project funds will realize, while the modified ROR technique requires this same information, plus
the borrowing rate for the organization considering the project. Usually, the investment rate is set
equal to the MARR, and the borrowing rate takes on the historical WACC rate. Each technique
will result in slightly different rates, but they are reliable for making the economic decision,
whereas the multiple rates are often not useful to decision making.
If an exact ROR is not necessary, it is strongly recommended that the PW or AW method at
the MARR be used to decide upon economic justification.
PROBLEMS
Understanding ROR
7.1 Under what circumstances would the rate of return
be ( a ) 100%, and ( b ) infinite?
7.2 A shrewd investor loaned $1,000,000 to a start-up
company at 10% per year interest for 3 years, but
the terms of the agreement were such that interest
would be charged on the principal rather than on
the unpaid balance. How much extra interest did
the company pay?
7.3 What is the nominal rate of return per year on an
investment that increases in value by 8% every
3 months?
7.4 Assume you borrow $50,000 at 10% per year interest
and you agree to repay the loan in five equal
annual payments. What is the amount of the unrecovered
balance immediately after you make the
third payment?
7.5 International Potash got a $50 million loan amortized
over a 10-year period at 10% per year interest.
The loan agreement stipulates that the loan
will be repaid in 10 equal annual payments with
interest charged on the principal amount of the
loan (not on the unrecovered balance).
( a) What is the amount of each payment?
( b) What is the total amount of interest paid?
How does the total interest paid compare
with the principal of the loan?
Determination of ROR
7.6 In 2010, the city of Houston, Texas, collected
$24,112,054 in fines from motorists because of
traffic violations caught by red-light cameras. The
cost of operating the system was $8,432,372. The
net profit, that is, profit after operating costs, is
split equally (that is, 50% each) between the city
and the operator of the camera system. What will
be the rate of return over a 3-year period to the
contractor that paid for, installed, and operates the
system, if its initial cost was $9,000,000 and the
profit for each of the 3 years is the same as it was
in 2010?
7.7 P&G sold its prescription drug business to
Warner-Chilcott, Ltd. for $3.1 billion. If income
from product sales is $2 billion per year and net
profit is 20% of sales, what rate of return will
the company make over a 10-year planning
horizon?
7.8 Water damage from a major flood in a Midwestern
city resulted in damages estimated at
$108 million. As a result of the claimant payouts,
insurance companies raised homeowners' insurance
rates by an average of $59 per year for each
of the 160,000 households in the affected city. If
a 20-year study period is considered, what was
the rate of return on the $108 million paid by the
insurance companies?

194 Chapter 7 Rate of Return Analysis: One Project
7.9 Determine the rate of return for the cash flows shown
in the diagram. (If requested by your instructor,
show both hand and spreadsheet solutions.)
to 1,694,247 in 2015. If the increase were to occur
uniformly, what rate of increase would be required
each year to meet the goal?
0 1 2 3 4 5 6 7 8
$3000
$200
$200 $200
i =?
$7000
$90 $90 $90
Year
7.14 U.S. Census Bureau statistics show that the annual
earnings for persons with a high school diploma
are $35,220 versus $57,925 for someone with a
bachelor’s degree. If the cost of attending college
is assumed to be $30,000 per year for 4 years and
the forgone earnings during those years are assumed
to be $35,220 per year, what rate of return
does earning a bachelor’s degree represent? Use a
35-year study period. ( Hint : The investment in
years 1 through 4 is the cost of college plus the
foregone earnings, and the income in years 5
through 35 is the difference in income between a
high school diploma and a bachelor’s degree.)
7.10 The Office of Naval Research sponsors a contest
for college students to build underwater robots that
can perform a series of tasks without human intervention.
The University of Florida, with its Subju-
Gator robot, won the $7000 first prize (and serious
bragging rights) over 21 other universities. If the
team spent $2000 for parts (at time 0) and the project
took 2 years, what annual rate of return did the
team make?
7.11 For the cash flows shown, determine the rate of
return.
Year 0 1 2 3 4 5
Expense, $ 17,000 2,500 2,500 2,500 2,500 2,500
Revenue, $ 0 5,000 6,000 7,000 8,000 12,000
7.12 In an effort to avoid foreclosure proceedings on
struggling mortgage customers, Bank of America
proposed an allowance that a jobless customer
make no payment on their mortgage for up to
9 months. If the customer did not find a job within
that time period, they would have to sign over their
house to the bank. The bank would give them
$2000 for moving expenses.
Assume John and his family had a mortgage
payment of $2900 per month and he was not able
to find a job within the 9-month period. If the bank
saved $40,000 in foreclosure costs, what rate of
return per month did the bank make on the allowance?
Assume the first payment that was skipped
was due at the end of month 1 and the $40,000
foreclosure savings and $2000 moving expense
occurred at the end of the 9-month forbearance
period.
7.13 The Closing the Gaps initiative by the Texas
Higher Education Coordinating Board established
the goal of increasing the number of students in
higher education in Texas from 1,064,247 in 2000
7.15 The Ester Municipal Water Utility issued 20-year
bonds in the amount of $53 million for several
high-priority flood control improvement projects.
The bonds carried a 5.38% dividend rate with the
dividend payable annually. The U.S. economy was
in a recession at that time, so as part of the federal
stimulus program, the Utility gets a 35% reimbursement
on the dividend it pays.
(a) What is the effective dividend rate that the
Utility is paying on the bonds?
(b)
What is the total dollar amount the Utility will
save in dividends over the life of the bonds?
(c) What is the future worth in year 20 of the
dividend savings, if the interest rate is
6% per year?
7.16 A contract between BF Goodrich and the Steelworkers
Union of America called for the company
to spend $100 million in capital investment to keep
the facilities competitive. The contract also required
the company to provide buyout packages
for 400 workers. If the average buyout package is
$100,000 and the company is able to reduce costs
by $20 million per year, what rate of return will the
company make over a 10-year period? Assume all
of the company’s expenditures occur at time 0 and
the savings begin 1 year later.
7.17 Rubber sidewalks made from ground up tires are
said to be environmentally friendly and easier on
people’s knees. Rubbersidewalks, Inc. of Gardena,
California, manufactures the small rubberized
squares that are being installed where tree roots,
freezing weather, and snow removal have required
sidewalk replacement or major repairs every
3 years. The District of Columbia spent $60,000 for
a rubber sidewalk to replace broken concrete in a
residential neighborhood lined with towering willow
oaks. If a concrete sidewalk costs $28,000 and

Problems 195
lasts only 3 years versus a 9-year life for the rubber
sidewalks, what rate of return does this represent?
7.18 Efficient light jets (ELJs) are smaller aircraft that
may revolutionize the way people travel by plane.
They cost between $1.5 and $3 million, seat 5 to 7
people, and can fly up to 1100 miles at cruising
speeds approaching 425 mph. Eclipse Aerospace
was founded in 2009, and its sole business is making
ELJs. The company invested $500 million
(time 0) and began taking orders 2 years later. If
the company accepted orders for 2500 planes and
received 10% down (in year 2) on planes having
an average cost of $1.8 million, what rate of return
will the company make over a 10-year planning
period? Assume 500 of the planes are delivered
each year in years 6 through 10 and that the company’s
M&O costs average $10 million per year in
years 1 through 10. (If requested by your instructor,
show both hand and spreadsheet solutions.)
7.19 Betson Enterprises distributes and markets the Big
Buck video game which allows players to “hunt”
for elk, antelope, moose, and bucks without shivering
outside in the cold. E-sports entertainment in
New York City purchased five machines for $6000
each and took in an average of $600 total per week
in sales. What rate of return does this represent
( a ) per week and ( b ) per year (nominal)? Use a
3-year study period with 52 weeks per year.
7.20 A 473-foot, 7000-ton World War II troop carrier
(once commissioned as the USS Excambion ) was
sunk in the Gulf of Mexico to serve as an underwater
habitat and diving destination. The project
took 10 years of planning and cost $4 million.
Assume the $4 million was expended equally in
years 1 through 10. What rate of return does the
venture represent, if increased fishing and recreation
activities are valued at $270,000 per year
beginning in year 11 and they continue in perpetuity?
(If assigned by your instructor, show both
hand and spreadsheet solutions.)
Multiple ROR Values
7.21 What is meant by a nonconventional cash flow
series?
7.22 Explain at least three types of projects in which
large net cash flow changes may cause sign changes
during the life of the project, thus indicating the
possible presence of multiple ROR values.
7.23 Explain a situation with which you are personally
familiar for which the net cash flows have
changed signs in a fashion similar to those in
Figure 7–5 .
7.24 According to Descartes’ rule of signs, what is the
maximum number of real-number values that will
balance a rate of return equation?
7.25 According to Descartes’ rule of signs, how many
possible i * values are there for net cash flows that
have the following signs?
(a)
(b)
(c)
7.26 According to Norstrom’s criterion, there are two
requirements regarding the cumulative cash flows
that must be satisfied to ensure that there is only
one positive root in a rate of return equation. What
are they?
7.27 According to Descartes’ rule of signs, how many
possible i * values are there for the cash flows
shown?
Year 1 2 3 4 5 6
Net Cash
Flow, $
4100 2000 7000 12,000 700 800
7.28 According to Descartes’ rule of signs, how many
i* values are possible for the cash flows shown?
Year 1 2 3 4
Revenue, $ 25,000 13,000 4,000 70,000
Costs, $ 30,000 7,000 6,000 12,000
7.29 According to Descartes’ rule and Norstrom’s criterion,
how many i* values are possible for the cash
flow (CF ) sequence shown?
Year 1 2 3 4 5
Net Cash 16,000 32,000 25,000 50,000 8,000
Flow, $
Cumulative
CF, $
16,000 16,000 41,000 9,000 1,000
7.30 For the cash flows shown, determine the sum of
the cumulative cash flows.
Year 0 1 2 3 4
Revenue, $ 25,000 15,000 4,000 18,000
Costs, $ 6,000 30,000 7,000 6,000 12,000
7.31 Stan-Rite Corp of Manitowoc, Wisconsin, is a B to
B company that manufactures many types of industrial
products, including portable measuring
arms with absolute encoders, designed to perform
3D inspections of industrial parts. If the company’s
cash flow (in millions) for one of its product divisions
is as shown on the next page, determine
( a ) the number of possible i * values and ( b ) all rate
of return values between 0% and 100%.

196 Chapter 7 Rate of Return Analysis: One Project
Year Expenses Revenues
0 $30 $ 0
1 20 18
2 25 19
3 15 36
4 22 52
5 20 38
6 30 70
7.32 Julie received a $50 bill for her birthday at the end
of January. At the end of February, she spent this
$50 and an additional $150 to buy clothes. Her
parents then gave her $50 and $125 at the end of
March and April, respectively, as she prepared to
go to summer school and needed the clothes. Her
conclusion was that over the 4 months, she had received
$25 more than she spent. Determine if Julie
has a multiple rate of return situation for these cash
flows. If so, determine the multiple rates and comment
on their validity. The cash flow values are as
follows:
Month Jan (0) Feb (1) Mar (2) Apr (3)
Cash Flow, $ 50 200 50 125
7.33 Veggie Burger Boy sells franchises to individuals
who want to start small in the sandwiches-forvegeterians
business and grow in net cash flow
over the years. A franchisee in Mississippi provided
the $5000 up-front money, but did very
poorly the first year. He was allowed to borrow at
the end of his first year from the corporation’s capital
incentive fund with a promise to repay the loan
in addition to the annual share that the corporation
contractually receives from annual sales. The net
cash flows from the corporation’s perspective are
shown below.
Year 0 1 2 3 4 5 6
NCF, $ 5000 10,100 500 2000 2000 2000 2000
The corporate chief financial officer (CFO) has
some questions concerning this NCF series. Help
her by doing the following, using a spreadsheet.
(a) Plot the PW versus i graph and estimate the
rate of return for this franchise.
(b)
Use the IRR function on a spreadsheet to find
the corresponding return.
(c) Basing your conclusions on Descartes’ and
Norstrom’s rules, provide the CFO with
some advice on what ROR value is the most
reliable for this franchise over the 6-year period.
The normal corporate MARR used for
franchisee evaluation is 30% per year.
7.34 In 2011, Vaught Industries closed its plant in
Marionsville following labor, environmental, and
safety problems. The plant was built in 2005
based on an older technology to produce meat
products. It had positive NCF until 2010 and discontinued
operation in 2011 due to labor and
safety problems. In 2012, prior to the sale of the
facility and property, Vaught spent $1 million to
make the site environmentally acceptable to a potential
buyer. The net cash flows in $100,000 over
the years are listed below. Use a spreadsheet to do
the following.
(a) Check for multiple rates of return.
(b) Find all rates that are real numbers between
25% and 50%, and calculate the PW
value for interest rates in this range.
(c) Indicate which is the best and correct i * value
to use in a PW analysis.
Year 2005 2006 2007 2008 2009 2010 2011 2012
NCF, $ 25 10 10 15 15 5 6 10
7.35 Five years ago, VistaCare spent $5 million to develop
and introduce a new service in home health
care for people who require frequent blood dialysis
treatments. The service was not well received
after the first year and was removed from the market.
When reintroduced 4 years after its initial
launch, it was much more profitable. Now, in year
5, VistaCare has spent a large sum on research to
broaden the application of this service. Use the
NCF series below to plot the PW versus i graph
and estimate the ROR over the 5 years. NCF values
are in $1 million units.
Year 0 1 2 3 4 5
Net Cash Flow, $ 5000 5000 0 0 15,000 15,000
Removing Multiple i * Values
7.36 In calculating the external rate of return by the
modified rate of return approach, it is necessary to
use two different rates of return, the investment rate
i i and the borrowing rate i b . When is each used?
7.37 In the modified rate of return approach for determining
a single interest rate from net cash flows,
state which interest rate is usually higher, the investment
rate i i or the borrowing rate i b . State why.
7.38 Use the modified rate of return approach with an
investment rate of 18% per year and a borrowing
rate of 10% to find the external rate of return for
the following cash flows.
Year 0 1 2 3
Net Cash Flow, $ 16,000 32,000 25,000 70,000
7.39 Harley worked for many years to save enough money
to start his own residential landscape design business.
The cash flows shown are those he recorded for the
first 6 years as his own boss. Find the external rate of

Problems 197
return using the modified rate of return approach, an
investment rate of 15% per year, and a borrowing rate
of 8%. (After using the procedure, use the MIRR
function to confirm your answer.)
Year 0 1 2 3 4 5 6
NCF, $ 9000 4100 2000 7000 12,000 700 800
7.40 Samara, an engineer working for GE, invested her
bonus money each year in company stock. Her
bonus has been $8000 each year for the past
6 years (i.e., at the end of years 1 to 6). At the end
of year 7, she sold the stock for $52,000 to buy a
condo; she purchased no stock that year. In years 8
to 10, she again invested the $8000 bonus. Samara
sold all of the remaining stock for $28,000 immediately
after the investment at the end of year 10.
( a) Determine the number of possible rate of return
values in the net cash flow series using
the two sign tests.
( b) Determine the external rate of return by
hand, using the modified rate of return approach
with an investment rate of 12% per
year and a borrowing rate of 8%.
( c) Find the external rate of return by spreadsheet
using the ROIC approach with an investment
rate of 12% per year.
(d) Enter the cash flows into a spreadsheet, and use
the IRR function to find the i * value. You should
get the same value as the ROIC in part ( c ). Explain
why this is so, given that the investment
rate is 12% per year. ( Hint : Look carefully at the
column labeled “Future worth, F, $” when you
solved part ( c ) using the spreadsheet.)
7.41 Swagelok Co. of Solon, Ohio, makes variable area
flowmeters (VAFs) that measure liquid and gas flow
rates by means of a tapered tube and float. If tooling
and setup costs were $400,000 in year 0 and an additional
$190,000 in year 3, determine the external
rate of return using the modified rate of return approach.
The revenue was $160,000 per year in
years 1 through 10. Assume the company’s MARR
is 20% per year and its cost of capital is 9% per year.
7.42 A company that makes clutch disks for race cars
has the cash flows shown for one department.
Year Cash Flow, $1000
0 65
1 30
2 84
3 10
4 12
( a) Determine the number of positive roots to the
rate of return relation.
( b) Calculate the internal rate of return.
(c)
Calculate the external rate of return using the
return on invested capital (ROIC) approach
with an investment rate of 15% per year. (As
assigned by your instructor, solve by hand
andor spreadsheet.)
7.43 For the cash flow series below, calculate the external
rate of return, using the return on invested capital
approach with an investment rate of 14% per
year.
Year Cash Flow, $
0 3000
1 2000
2 1000
3 6000
4 3800
7.44 Five years ago, a company made a $500,000 investment
in a new high-temperature material. The
product did poorly after only 1 year on the market.
However, with a new name and advertising campaign
4 years later it did much better. New development
funds have been expended this year
(year 5) at a cost of $1.5 million. Determine the
external rate of return using the ROIC approach
and an investment rate of 15% per year. The i * rate
is 44.1% per year.
Year Cash Flow, $
0 500,000
1 400,000
2 0
3 0
4 2,000,000
5 1,500,000
Bonds
7.45 What is the bond coupon rate on a $25,000 mortgage
bond that has semiannual interest payments
of $1250 and a 20-year maturity date?
7.46 An equipment trust bond with a face value of
$10,000 has a bond coupon rate of 8% per year,
payable quarterly. What are the amount and frequency
of the dividend payments?
7.47 What is the face value of a municipal bond that
matures in 20 years and has a bond coupon rate
of 6% per year with semiannual payments of
$900?
7.48 What is the present worth of a $50,000 debenture
bond that has a bond coupon rate of 8% per year,
payable quarterly? The bond matures in 15 years.
The interest rate in the marketplace is 6% per year,
compounded quarterly.

198 Chapter 7 Rate of Return Analysis: One Project
7.49 Best Buy issued collateral bonds 4 years ago that
have a face value of $20,000 each and a coupon
rate of 8% per year, payable semiannually. If the
bond maturity date is 20 years from the date they
were issued and the interest rate in the marketplace
is now 12% per year, compounded semiannually,
what is the present worth (now) of one bond?
7.50 In 2011, El Paso Water Utilities (EPWU) issued
bonds worth $9.125 million to improve the Van
Buren dam in central El Paso and to finance three
other drainage projects. The bonds were purchased
by the Texas Water Development Board under the
federal stimulus program wherein EPWU did not
have to pay any dividend on the bonds. If the bond
dividend rate would have been 4% per year, payable
quarterly, with a bond maturity date 18 years
after issuance, what is the present worth of the
dividend savings to EPWU rate payers? Assume
the market interest rate is 6% per year.
7.51 A recently issued industrial bond with a face value
of $10,000 has a coupon rate of 8% per year, payable
annually. The bond matures 20 years from now. Jeremy
is interested in buying one bond. If he pays
$10,000 for the bond and plans to hold it to maturity,
what rate of return per year will he realize?
7.52 Due to a significant troop buildup at the local military
base, a school district issued $10,000,000 in
bonds to build new schools. The bond coupon rate
is 6% per year, payable semiannually, with a maturity
date of 20 years. If an investor is able to purchase
one of the bonds that has a face value of
$5000 for $4800, what rate of return per 6 months
will the investor realize? Assume the bond is kept
to maturity.
7.53 As the name implies, a zero-coupon bond pays no
dividend, only the face value when it matures. If a
zero coupon bond that has a face value of $10,000
and a maturity date of 15 years is for sale for
$2000, what rate of return will the purchaser make,
provided the bond is held to maturity?
7.54 To provide infrastructure in the outlying areas of
Morgantown, West Virginia, the city council
issued 30-year bonds with a face value of $25 million.
The bond coupon rate was set at 5% per year,
payable semiannually. Because the market interest
rate increased immediately before the bonds were
sold, the city received only $23.5 million from the
bond sale. What was the semiannual interest rate
when the bonds were sold?
7.55 An investor who purchased a $10,000 mortgage
bond today paid only $6000 for it. The bond coupon
rate is 8% per year, payable quarterly, and the
maturity date is 18 years from the year of issuance.
Because the bond is in default, it will pay no dividend
for the next 2 years. If the bond dividend is in
fact paid for the following 5 years (after the
2 years) and the investor then sells the bond for
$7000, what rate of return will be realized ( a ) per
quarter and ( b ) per year (nominal)?
7.56 Five years ago, GSI, an oil services company
headquartered in Texas, issued $10 million worth
of 12% 30-year bonds with the dividend payable
quarterly. The bonds have a call date of this year if
GSI decides to take advantage of it. The interest
rate in the marketplace decreased enough that the
company is considering calling the bonds since the
coupon rate is relatively high. If the company buys
the bonds back now for $11 million, determine the
rate of return that the company will make ( a ) per
quarter and ( b ) per year (nominal). ( Hint : The
“call” option means the following: By spending
$11 million now, the company will not have to
make the quarterly bond dividend payments or pay
the face value of the bonds when they come due
25 years from now.)
7.57 Four years ago, Chevron issued $5 million worth
of debenture bonds with a coupon rate of 10% per
year, payable semiannually. Market interest rates
dropped, and the company called the bonds (i.e.,
paid them off in advance) at a 10% premium on the
face value. Therefore, it cost the corporation
$5.5 million to retire the bonds. What semiannual
rate of return did an investor make who purchased
a $5000 bond 4 years ago and held it until it was
called 4 years later?
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
7.58 All of the following mean the same as rate of
return except:
(a) Internal rate of return
(b) Time for return of capital
(c) Interest rate
(d) Return on investment
7.59 The numerical value of i in a rate of return equation
can range from:
(a) 0% to 100%
(b) 0 to
(c) 100% to 100%
(d) 100% to

Additional Problems and FE Exam Review Questions 199
7.60 The internal rate of return on an investment refers
to the interest rate earned on the:
( a) Initial investment
( b) Unrecovered balance of the investment
( c) Money recovered from an investment
( d) Income from an investment
7.61 A conventional (or simple) cash flow series is one
wherein :
( a) The algebraic signs on the net cash flows
change only once.
( b) The interest rate you get is a simple interest
rate.
( c) The total of the net cash flows is equal to 0.
( d) The total of the cumulative cash flows is
equal to 0.
7.62 According to Descartes’ rule of signs, for a net
cash flow sequence of , the number of
possible i values is:
( a) 2
( b) 3
( c) 4
( d) 5
7.63 According to Norstrom’s criterion, the one statement
below that is correct is :
( a) The cumulative cash flow must start out
positively.
( b) The cumulative cash flow must start out
negatively.
( c) The cumulative cash flow must equal 0.
( d) The net cash flow must start out positively.
7.64 According to Descartes’ rule and Norstrom’s criterion,
the number of positive i * values for the
following cash flow sequence is:
Year 1 2 3 4
Revenue, $ 25,000 15,000 4,000 18,000
Cost, $ 30,000 7,000 6,000 12,000
( a) 1
( b) 2
( c) 3
( d) 4
7.65 For the net cash flows and cumulative cash flows
shown, the value of x is:
Year 1 2 3 4 5
NCF, $ 13,000 29,000 25,000 x 8000
Cumulative 13,000 16,000 41,000 9000 1000
NCF, $
( a) $7000
( b) $16,000
( c) $36,000
( d) $50,000
7.66 A company that uses a minimum attractive rate of
return of 10% per year is evaluating new processes
to improve operational efficiency. The estimates
associated with candidate processes are shown.
Alternative I
Alternative J
First cost, $ 40,000 50,000
Annual cost, $ per year 15,000 12,000
Salvage value, $ 5,000 5,000
Life, years 3 6
The statement that is most correct is :
(a) The alternatives are revenue alternatives.
(b) The alternatives are cost alternatives.
(c) The alternatives are revenue alternatives and
DN is an option.
(d) The alternatives are cost alternatives and DN
is an option.
7.67 Scientific Instruments, Inc. uses a MARR of 8% per
year. The company is evaluating a new process to
reduce water effluents from its manufacturing processes.
The estimate associated with the process follows.
In evaluating the process on the basis of a rate
of return analysis, the correct equation to use is:
(a)
(b)
(c)
(d)
New Process
First cost, $ 40,000
NCF, $ per year 13,000
Salvage value, $ 5,000
Life, years 3
0 40,000 13,000( P A , i ,3)
5000( P F , i ,3)
0 40,000( AP,i ,3) 13,000
5000( A F , i ,3)
0 40,000( FP,i ,3) 13,000( FA,i, 3)
5000
Any of the above
7.68 When one is using the modified ROR method to remove
multiple ROR values, an additional estimate
needed besides the cash flows and their timings is:
(a) The ROIC value
(b) External rate of return
(c) Investment rate
(d) Internal rate of return
7.69 For the following cash flows, the modified rate of
return method uses a borrowing rate of 10%, and
an investment rate is 12% per year. The correct
computation for the present worth in year 0 is:
Year 1 2 3 4 5
NCF, $ 10,000 0 0 19,000 25,000
(a) 10,000 19,000( P F ,12%,4)
(b) 10,000 19,000( P F ,12%,4)
25,000( P F ,10%,5)

200 Chapter 7 Rate of Return Analysis: One Project
(c) 25,000( P F ,10%,5)
(d) 10,000 19,000( P F ,10%,4)
7.70 The return on invested capital (ROIC) method removes
multiple ROR values from a cash flow sequence.
If the future worth computation in year t is
F t 0, the ROIC rate i is used. The interpretation
of F t 0 most closely means:
(a) The net balance of project cash flows in year
t is negative.
(b) The resulting external rate of return will be
(c)
positive.
The net balance of project cash flows is positive
in year t .
(d) The sequence has nonremovable negative
ROR values.
7.71 The meaning of return on invested capital for a
corporation is best stated as :
(a) A rate-of-return measure that equates the internal
and external ROR
(b) A measure of how effectively the corporation
uses capital funds invested in it
(c) The value at which borrowing ROR and investing
ROR are equal
(d) The external rate of return value is based on
total capital invested
7.72 A corporate $10,000 bond has a coupon rate of 8%
per year, payable semiannually. The bond matures
20 years from now. At an interest rate of 6% per
year, compounded semiannually, the amount and
frequency of the bond dividend payments are :
(a) $600 every 6 months
(b) $800 every 6 months
(c) $300 every 6 months
(d) $400 every 6 months
7.73 A $20,000 mortgage bond that is due in 1 year
pays interest of $500 every 3 months. The bond’s
coupon rate is:
(a) 2.5% per year, payable quarterly
(b) 5% per year, payable quarterly
(c) 5% per year, payable semiannually
(d) 10% per year, payable quarterly
7.74 A $10,000 municipal bond due in 10 years pays
interest of $400 every 6 months. If an investor purchases
the bond now for $9000 and holds it to maturity,
the rate of return received can be determined
by the following equation:
(a) 0 9000 400( P A , i ,10)
10,000( P F , i ,10)
(b) 0 9000 400( P A , i ,20)
10,000( P F , i ,20)
(c) 0 10,000 400( P A , i ,20)
10,000( P F , i ,20)
(d) 0 9000 800( P A , i ,10)
10,000( P F , i ,10)
7.75 A debenture bond issued 3 years ago has a face
value of $5000, a coupon rate of 4% per year, payable
semiannually, and a maturity date of 20 years
from the date it was issued . The bond is for sale
now for $4500. If the interest rate in the marketplace
is compounded quarterly, the value of n that
must be used in the P A factor to calculate the rate
of return for the bond is:
( a) 34
( b) 40
( c) 68
( d) 80
CASE STUDY
DEVELOPING AND SELLING AN INNOVATIVE IDEA
Background
Three engineers who worked for Mitchell Engineering, a
company specializing in public housing development, went
to lunch together several times a week. Over time they decided
to work on solar energy production ideas. After a lot of
weekend time over several years, they had designed and developed
a prototype of a low-cost, scalable solar energy plant
for use in multifamily dwellings on the low end and mediumsized
manufacturing facilities on the upper end. For residential
applications, the collector could be mounted along side a
TV dish and be programmed to track the sun. The generator
and additional equipment are installed in a closet-sized area
in an apartment or on a floor for multiple-apartment supply.
The system serves as a supplement to the electricity provided
by the local power company. After some 6 months of testing,
it was agreed that the system was ready to market and reliably
state that an electricity bill in high-rises could be reduced by
approximately 40% per month. This was great news for lowincome
dwellers on government subsidy that are required to
pay their own utility bills.
Information
With a hefty bank loan and $200,000 of their own capital,
they were able to install demonstration sites in three cities in
the sunbelt. Net cash flow after all expenses, loan repayment,
and taxes for the first 4 years was acceptable; $55,000 at the

Case Study 201
end of the first year, increasing by 5% each year thereafter. A
business acquaintance introduced them to a potential buyer of
the patent rights and current subscriber base with an estimated
$500,000 net cashout after only these 4 years of ownership.
However, after serious discussion replaced the initial
excitement of the sales offer, the trio decided to not sell at this
time. They wanted to stay in the business for a while longer
to develop some enhancement ideas and to see how much
revenue may increase over the next few years.
During the next year, the fifth year of the partnership, the
engineer who had received the patents upon which the collector
and generator designs were based became very displeased
with the partnering arrangements and left the trio to go into
partnership with an international firm in the energy business.
With new research and development funds and the patent
rights, a competing design was soon on the market and took
much of the business away from the original two developers.
Net cash flow dropped to $40,000 in year 5 and continued to
decrease by $5000 per year. Another offer to sell in year 8
was presented, but it was only for $100,000 net cash. This
was considered too much of a loss, so the two owners did not
accept. Instead, they decided to put $200,000 more of their
own savings into the company to develop additional applications
in the housing market.
It is now 12 years since the system was publicly launched.
With increased advertising and development, net cash flow
has been positive the last 4 years, starting at $5000 in year 9
and increasing by $5000 each year until now.
Case Study Exercises
It is now 12 years after the products were developed, and the
engineers invested most of their savings in an innovative
idea. However, the question of “When do we sell?” is always
present in these situations. To help with the analysis, determine
the following:
1. The rate of return at the end of year 4 for two situations:
( a ) The business is sold for the net cash amount of
$500,000 and ( b ) no sale.
2. The rate of return at the end of year 8 for two situations:
( a ) The business is sold for the net cash amount of
$100,000 and ( b ) no sale.
3. The rate of return now at the end of year 12.
4. Consider the cash flow series over the 12 years. Is there
any indication that multiple rates of return may be present?
If so, use the spreadsheet already developed to
search for ROR values in the range 100% other than
the one determined in exercise 3 above.
5. Assume you are an investor with a large amount of
ready cash, looking for an innovative solar energy product.
What amount would you be willing to offer for the
business at this point (end of year 12) if you require a
12% per year return on all your investments and, if purchased,
you plan to own the business for 12 additional
years? To help make the decision, assume the current
NCF series continues increasing at $5000 per year for
the years you would own it. Explain your logic for
offering this amount.

CHAPTER 8
Rate of Return
Analysis:
Multiple
Alternatives
LEARNING OUTCOMES
Purpose: Select the best alternative on the basis of incremental rate of return analysis.
SECTION TOPIC LEARNING OUTCOME
8.1 Incremental analysis • State why the ROR method of comparing
alternatives requires an incremental cash flow
analysis.
8.2 Incremental cash flows • Calculate the incremental cash flow series for two
alternatives.
8.3 Meaning of i * • Interpret the meaning of the incremental ROR ( i *)
determined from the incremental cash flow series.
8.4 i * from PW relation • Based on a PW relation, select the better of two
alternatives using incremental ROR analysis or a
breakeven ROR value.
8.5 i * from AW relation • Select the better of two alternatives using
incremental ROR analysis based on an AW
relation.
8.6 More than two alternatives • Select the best from several alternatives using
incremental ROR analysis.
8.7 All-in-one spreadsheet • Use a single spreadsheet to perform PW, AW, ROR,
and incremental ROR analyses for mutually
exclusive and independent alternatives.

T
his chapter presents the methods by which two or more alternatives can be evaluated
using a rate of return (ROR) comparison based on the methods of the previous
chapter. The ROR evaluation, correctly performed, will result in the same selection
as the PW and AW analyses, but the computational procedure is considerably different for ROR
evaluations. The ROR analysis evaluates the increments between two alternatives in pairwise
comparisons. As the cash flow series becomes more complex, spreadsheet functions help speed
computations.
8.1 Why Incremental Analysis Is Necessary
When two or more mutually exclusive alternatives are evaluated, engineering economy can identify
the one alternative that is the best economically. As we have learned, the PW and AW techniques
can be used to do so, and are the recommended methods. Now the procedure using ROR
to identify the best is presented.
Let’s assume that a company uses a MARR of 16% per year, that the company has $90,000
available for investment, and that two alternatives (A and B) are being evaluated. Alternative
A requires an investment of $50,000 and has an internal rate of return i * A of 35% per year. Alternative
B requires $85,000 and has an i * B of 29% per year. Intuitively we may conclude that
the better alternative is the one that has the larger return, A in this case. However, this is not
necessarily so. While A has the higher projected return, its initial investment ($50,000) is much
less than the total money available ($90,000). What happens to the investment capital that is
left over? It is generally assumed that excess funds will be invested at the company’s MARR,
as we learned in previous chapters. Using this assumption, it is possible to determine the consequences
of the two alternative investments. If alternative A is selected, $50,000 will return
35% per year. The $40,000 left over will be invested at the MARR of 16% per year. The rate
of return on the total capital available, then, will be the weighted average. Thus, if alternative
A is selected,
50,000(0.35) 40,000(0.16)
Overall ROR A ———————————— 26.6%
90,000
If alternative B is selected, $85,000 will be invested at 29% per year, and the remaining $5000
will earn 16% per year. Now the weighted average is
85,000(0.29) 5000(0.16)
Overall ROR B ——————————— 28.3%
90,000
These calculations show that even though the i * for alternative A is higher, alternative B presents
the better overall ROR for the $90,000. If either a PW or AW comparison is conducted using the
MARR of 16% per year as i , alternative B will be chosen.
This simple example illustrates a major fact about the rate of return method for ranking and
comparing alternatives:
Under some circumstances, project ROR values do not provide the same ranking of alternatives
as do PW and AW analyses. This situation does not occur if we conduct an incremental ROR
analysis (discussed below).
When independent projects are evaluated, no incremental analysis is necessary between
projects. Each project is evaluated separately from others, and more than one can be selected.
Therefore, the only comparison is with the do-nothing alternative for each project. The project
ROR can be used to accept or reject each one.
Independent project
selection
8.2 Calculation of Incremental Cash Flows for
ROR Analysis
To conduct an incremental ROR analysis, it is necessary to calculate the incremental cash flow
series over the lives of the alternatives. Based upon the equivalence relations (PW and AW), ROR
evaluation makes the equal-service assumption.

204 Chapter 8 Rate of Return Analysis: Multiple Alternatives
TABLE 8–1
Year
0
1
Format for Incremental Cash Flow Tabulation
Cash Flow
Alternative A
(1)
Alternative B
(2)
Incremental
Cash Flow
(3) (2) (1)
Equal-service requirement
The incremental ROR method requires that the equal-service requirement be met. Therefore,
the LCM (least common multiple) of lives for each pairwise comparison must be used. All
the assumptions of equal service present for PW analysis are necessary for the incremental
ROR method.
A format for hand or spreadsheet solutions is helpful (Table 8–1). Equal-life alternatives have
n years of incremental cash flows, while unequal-life alternatives require the LCM of lives for
analysis. At the end of each life cycle, the salvage value and initial investment for the next cycle
must be included for the LCM case.
When a study period is established, only this number of years is used for the evaluation. All
incremental cash flows outside the period are neglected. As we learned earlier, using a study period,
especially one shorter than the life of either alternative, can change the economic decision
from that rendered when the full lives are considered.
Only for the purpose of simplification, use the convention that between two alternatives, the
one with the larger initial investment will be regarded as alternative B . Then, for each year in
Table 8–1,
Incremental cash flow cash flow B cash flow A [8.1]
The initial investment and annual cash flows for each alternative (excluding the salvage value)
are one of the types identified in Chapter 5:
Revenue alternative , where there are both negative and positive cash flows
Cost alternative, where all cash flow estimates are negative
Revenue or cost
alternative
In either case, Equation [8.1] is used to determine the incremental cash flow series with the sign
of each cash flow carefully determined.
EXAMPLE 8.1
A tool and die company in Hanover is considering the purchase of a drill press with fuzzy-logic
software to improve accuracy and reduce tool wear. The company has the opportunity to buy a
slightly used machine for $15,000 or a new one for $21,000. Because the new machine is a
more sophisticated model, its operating cost is expected to be $7000 per year, while the used
machine is expected to require $8200 per year. Each machine is expected to have a 25-year life
with a 5% salvage value. Tabulate the incremental cash flow.
Solution
Incremental cash flow is tabulated in Table 8–2. The subtraction performed is (new – used)
since the new machine has a larger initial cost. The salvage values in year 25 are separated
from ordinary cash flow for clarity. When disbursements are the same for a number of consecutive
years, for hand solution only , it saves time to make a single cash flow listing, as is
done for years 1 to 25. However, remember that several years were combined when performing
the analysis. This approach cannot be used for spreadsheets, when the IRR or NPV function is
used, as each year must be entered separately.

8.2 Calculation of Incremental Cash Flows for ROR Analysis 205
TABLE 8–2 Cash Flow Tabulation for Example 8.1
Year
Cash Flow
Used Press New Press
Incremental
Cash Flow
(New – Used)
0 $15,000 $21,000 $6,000
1–25 8,200 7,000 1,200
25 750 1,050 300
EXAMPLE 8.2
A sole-source vendor can supply a new industrial park with large transformers suitable for
underground utilities and vault-type installation. Type A has an initial cost of $70,000 and
a life of 8 years. Type B has an initial cost of $95,000 and a life expectancy of 12 years.
The annual operating cost for type A is expected to be $9000, while the AOC for type B is
expected to be $7000. If the salvage values are $5000 and $10,000 for type A and type B,
respectively, tabulate the incremental cash flow using their LCM for hand and spreadsheet
solutions.
Solution by Hand
The LCM of 8 and 12 is 24 years. In the incremental cash flow tabulation for 24 years
(Table 8–3), note that the reinvestment and salvage values are shown in years 8 and 16 for type
A and in year 12 for type B.
Solution by Spreadsheet
Figure 8–1 shows the incremental cash flows for the LCM of 24 years. As in the hand tabulation,
reinvestment is made in the last year of each intermediate life cycle. The incremental
values in column D are the result of subtractions of column B from C.
Note that the final row includes a summation check. The total incremental cash flow should
agree in both the column D total and the subtraction C29 B29. Also note that the incremental
values change signs three times, indicating the possibility of multiple i* values, per Descartes’
rule of signs. This possible dilemma is discussed later in the chapter.
TABLE 8–3 Incremental Cash Flow Tabulation, Example 8.2
Cash Flow
Incremental
Cash Flow
Year
Type A
Type B
(B A)
0 $ 70,000 $ 95,000 $25,000
1–7 9,000 –7,000 2,000
70,000
8 9,000 –7,000 67,000
5,000
9–11 9,000 –7,000 2,000
–95,000
12 9,000 –7,000 83,000
10,000
13–15 9,000 –7,000 2,000
70,000
16 9,000 –7,000 67,000
5,000
17–23 9,000 –7,000 2,000
24
9,000 –7,000
5,000 10,000
7,000
$411,000 $338,000 $73,000

206 Chapter 8 Rate of Return Analysis: Multiple Alternatives
Figure 8–1
Spreadsheet computation
of incremental
cash flows for
unequal-life alternatives,
Example 8.2.
Starting new life cycle for A
initial cost AOC salvage
70,000 9,000 5,000
Check on summations
Incremental column should
equal difference of columns
8.3 Interpretation of Rate of Return
on the Extra Investment
The incremental cash flows in year 0 of Tables 8–2 and 8–3 reflect the extra investment or cost
required if the alternative with the larger first cost is selected. This is important in an incremental
ROR analysis in order to determine the ROR earned on the extra funds expended for the largerinvestment
alternative. If the incremental cash flows of the larger investment don’t justify it, we
must select the cheaper one. In Example 8.1 the new drill press requires an extra investment of
$6000 (Table 8–2). If the new machine is purchased, there will be a “savings” of $1200 per year
for 25 years, plus an extra $300 in year 25. The decision to buy the used or new machine can
be made on the basis of the profitability of investing the extra $6000 in the new machine. If the
equivalent worth of the savings is greater than the equivalent worth of the extra investment at the
MARR, the extra investment should be made (i.e., the larger first-cost proposal should
be accepted). On the other hand, if the extra investment is not justified by the savings, select the
lower-investment proposal.
It is important to recognize that the rationale for making the selection decision is the same as
if only one alternative were under consideration, that alternative being the one represented by the
incremental cash flow series. When viewed in this manner, it is obvious that unless this investment
yields a rate of return equal to or greater than the MARR, the extra investment should not
be made. As further clarification of this extra investment rationale, consider the following: The
rate of return attainable through the incremental cash flow is an alternative to investing at the
MARR. Section 8.1 states that any excess funds not invested in the alternative are assumed to
be invested at the MARR. The conclusion is clear:
ME alternative selection
If the rate of return available through the incremental cash flow equals or exceeds the MARR,
the alternative associated with the extra investment should be selected.

8.4 Rate of Return Evaluation Using PW: Incremental and Breakeven 207
Not only must the return on the extra investment meet or exceed the MARR, but also the return
on the investment that is common to both alternatives must meet or exceed the MARR. Accordingly,
prior to performing an incremental ROR analysis, it is advisable to determine the internal rate of
return i * for each alternative. This can be done only for revenue alternatives, because cost alternatives
have only cost (negative) cash flows and no i * can be determined. The guideline is as follows:
For multiple revenue alternatives, calculate the internal rate of return i * for each alternative,
and eliminate all alternatives that have an i * MARR. Compare the remaining alternatives
incrementally.
As an illustration, if the MARR 15% and two alternatives have i * values of 12% and 21%,
the 12% alternative can be eliminated from further consideration. With only two alternatives, it
is obvious that the second one is selected. If both alternatives have i * MARR, no alternative is
justified and the do-nothing alternative is the best economically. When three or more alternatives
are evaluated, it is usually worthwhile, but not required, to calculate i * for each alternative for
preliminary screening. Alternatives that cannot meet the MARR may be eliminated from further
evaluation using this option. This option is especially useful when performing the analysis by
spreadsheet. The IRR function applied to each alternative’s cash flow estimates can quickly indicate
unacceptable alternatives, as demonstrated in Section 8.6.
When independent projects are evaluated, there is no comparison on the extra investment.
The ROR value is used to accept all projects with i * MARR, assuming there is no budget
limitation. For example, assume MARR 10%, and three independent projects are available
with ROR values of
Independent project
selection
i A
* 12%
i * B 9% i C
* 23%
Projects A and C are selected, but B is not because i * B MARR.
8.4 Rate of Return Evaluation Using PW:
Incremental and Breakeven
In this section we discuss the primary approach to making mutually exclusive alternative selections
by the incremental ROR method. A PW-based relation is developed for the incremental cash
flows and set equal to zero. Use hand solution or spreadsheet functions to find i * B–A , the internal
ROR for the series. Placing (delta) before i * B–A distinguishes it from the overall ROR values i A
*
and i B
*. ( i * may replace i * B–A when only two alternatives are present.)
Since incremental ROR requires equal-service comparison, the LCM of lives must be used in
the PW formulation. Because of the reinvestment requirement for PW analysis for different-life
assets, the incremental cash flow series may contain several sign changes, indicating multiple i *
values. Though incorrect, this indication is usually neglected in actual practice. The correct
approach is to follow one of the techniques of Section 7.5. This means that the single external
ROR ( i or i ) for the incremental cash flow series is determined.
These three elements— incremental cash flow series, LCM, and multiple roots —are the primary
reasons that the ROR method is often applied incorrectly in engineering economy analyses of
multiple alternatives. As stated earlier, it is always possible, and generally advisable, to use a PW or
AW analysis at an established MARR in lieu of the ROR method when multiple rates are indicated.
The complete procedure for hand or spreadsheet solution for an incremental ROR analysis
for two alternatives is as follows:
1. Order the alternatives by initial investment or cost, starting with the smaller one, called
A. The one with the larger initial investment is in the column labeled B in Table 8–1.
2. Develop the cash flow and incremental cash flow series using the LCM of years, assuming
reinvestment in alternatives.
3. Draw an incremental cash flow diagram, if needed.

208 Chapter 8 Rate of Return Analysis: Multiple Alternatives
ME alternative
selection
4. Count the number of sign changes in the incremental cash flow series to determine if
multiple rates of return may be present. If necessary, use Norstrom’s criterion to determine
if a single positive root exists.
5. Set up the PW 0 equation and determine i* B–A .
6. Select the economically better alternative as follows:
If i* BA MARR, select alternative A.
If i* BA MARR, the extra investment is justified; select alternative B.
If i* is exactly equal to or very near the MARR, noneconomic considerations help in
the selection of the “better” alternative.
EXAMPLE 8.3
In step 5, if trial and error by hand is used, time may be saved if the i * BA value is bracketed,
rather than approximated by a point value using linear interpolation, provided that a single ROR
value is not needed. For example, if the MARR is 15% per year and you have established that
i * BA is in the 15% to 20% range, an exact value is not necessary to accept B since you already
know that i * BA MARR.
The IRR function on a spreadsheet will normally determine one i * value. Multiple guess
values can be input to find multiple roots in the range 100% to for a nonconventional series,
as illustrated in Example 7.4. If this is not the case, to be correct, the indication of multiple roots
in step 4 requires that one of the techniques of Section 7.5 be applied to find an EROR.
As Ford Motor Company retools an old truck assembly plant in Michigan to produce a fuelefficient
economy model, the Ford Focus. Ford and its suppliers are seeking additional sources for
light, long-life transmissions. Automatic transmission component manufacturers use highly finished
dies for precision forming of internal gears and other moving parts. Two United States–
based vendors make the required dies. Use the per unit estimates below and a MARR of 12% per
year to select the more economical vendor bid. Show both hand and spreadsheet solutions.
Initial cost, $ 8,000 13,000
Annual costs, $ per year 3,500 1,600
Salvage value, $ 0 2,000
Life, years 10 5
Solution by Hand
These are cost alternatives, since all cash flows are costs. Use the procedure described above to
determine i * B–A .
1. Alternatives A and B are correctly ordered with the higher first-cost alternative in column 2
of Table 8–4.
2. The cash flows for the LCM of 10 years are tabulated.
TABLE 8–4 Incremental Cash Flow Tabulation, Example 8.3
Year
Cash Flow A
(1)
A
Cash Flow B
(2)
B
Incremental
Cash Flow
(3) (2) (1)
0 $ 8,000 $13,000 $ 5,000
1–5 3,500 1,600 1,900
5 —
2,000
13,000
11,000
6–10 3,500 1,600 1,900
10 — 2,000 2,000
$43,000 $38,000 $ 5,000

8.4 Rate of Return Evaluation Using PW: Incremental and Breakeven 209
$1900
$2000
Figure 8–2
Diagram of incremental cash
flows, Example 8.3 .
0 1 2 3 4 5 6 7 8 9
10
$5000
$11,000
3. The incremental cash flow diagram is shown in Figure 8–2 .
4. There are three sign changes in the incremental cash flow series, indicating as many as
three roots. There are also three sign changes in the cumulative incremental series, which
starts negatively at S 0 $5000 and continues to S 10 $5000, indicating that more
than one positive root may exist.
5. The rate of return equation based on the PW of incremental cash flows is
0 5000 1900( P A , i *,10) 11,000( P F , i *,5) 2000( P F , i *,10) [8.2]
In order to resolve any multiple-root problem, we can assume that the investment rate i i in the
ROIC technique will equal the i* found by trial and error. Solution of Equation [8.2] for the
first root discovered results in i * between 12% and 15%. By interpolation i * 12.65%.
6. Since the rate of return of 12.65% on the extra investment is greater than the 12% MARR,
the higher-cost vendor B is selected.
Comment
In step 4, the presence of up to three i * values is indicated. The preceding analysis finds one of
the roots at 12.65%. When we state that the incremental ROR is 12.65%, we assume that any
positive net cash flows are reinvested at 12.65%. If this is not a reasonable assumption, the
ROIC or modified ROR technique (Section 7.5) must be applied to find a different single i '
or i '' to compare with MARR 12%.
The other two roots are very large positive and negative numbers, as the IRR function of
Excel reveals. So they are not useful to the analysis.
Solution by Spreadsheet
Steps 1 through 4 are the same as above.
5. Figure 8–3 includes the same incremental net cash flows from Table 8–4 calculated in
column D. Cell D15 displays the i * value of 12.65% using the IRR function.
Figure 8–3
Spreadsheet solution using LCM of
lives and IRR function, Example 8.3 .
NPV(12%,D5:D14) + D4

210 Chapter 8 Rate of Return Analysis: Multiple Alternatives
6. Since the rate of return on the extra investment is greater than the 12% MARR, the highercost
vendor B is selected.
Comment
Once the spreadsheet is set up, there are a wide variety of analyses that can be performed. For
example, row 17 uses the NPV function to verify that the present worth is positive at MARR12%.
Charts such as PW versus i and PW versus i help graphically interpret the situation.
The rate of return determined for the incremental cash flow series or the actual cash flows can
be interpreted as a breakeven rate of return value.
The breakeven rate of return is the incremental i* value, i*, at which the PW (or AW)
value of the incremental cash fl ows is exactly zero. Equivalently, the breakeven ROR is the i
value, i*, at which the PW (or AW) values of two alternatives’ actual cash flows are exactly
equal to each other.
Breakeven ROR
If the incremental cash flow ROR ( i *) is greater than the MARR, the larger-investment
alternative is selected. For example, if the PW versus i graph for the incremental cash flows in
Table 8–4 (and spreadsheet Figure 8–3 ) is plotted for various interest rates, the graph shown in
Figure 8–4 is obtained. It shows the i * breakeven at 12.65%. The conclusions are that
• For MARR 12.65%, the extra investment for B is justified.
• For MARR 12.65%, the opposite is true—the extra investment in B should not be made,
and vendor A is selected.
• If MARR is exactly 12.65%, the alternatives are equally attractive.
Figure 8–5 , which is a breakeven graph of PW versus i for the cash flows (not incremental) of
each alternative in Example 8.3 , provides the same results. Since all net cash flows are negative
1800
1600
1400
1200
For MARR
in this range,
select B
Breakeven i
is 12.65%
For MARR
in this range,
select A
PW of incremental cash flows, $
1000
800
600
400
200
0
– 200
6 7 8 9
10 11 12
14 15
16 i%
– 400
– 600
– 800
Vendor B
Vendor A
Figure 8–4
Plot of present worth of incremental cash flows for Example 8.3 at various
i values.

8.4 Rate of Return Evaluation Using PW: Incremental and Breakeven 211
PW of alternative cash flows, $
25,000
26,000
27,000
28,000
29,000
30,000
Interest rate, % Figure 8–5
10 11 12 13 14 15 16
Breakeven graph of
Example 8.3 cash flows
(not incremental).
B
A
A
B
(cost alternatives), the PW values are negative. Now, the same conclusions are reached using the
following logic:
• If MARR 12.65%, select B since its PW of cost cash flows is smaller (numerically larger).
• If MARR 12.65%, select A since its PW of costs is smaller.
• If MARR is exactly 12.65%, either alternative is equally attractive.
Example 8.4 illustrates incremental ROR evaluation and breakeven rate of return graphs for
revenue alternatives. More of breakeven analysis is covered in Chapter 13.
EXAMPLE 8.4
New filtration systems for commercial airliners are available that use an electric field to remove up
to 99.9% of infectious diseases and pollutants from aircraft air. This is vitally important, as many
of the flu germs, viruses, and other contagious diseases are transmitted through the systems that
recirculate aircraft air many times per hour. Investments in the new filtration equipment can cost
from $100,000 to $150,000 per aircraft, but savings in fuel, customer complaints, legal actions,
etc., can also be sizable. Use the estimates below (in $100 units) from two suppliers provided to an
international carrier to do the following, using a spreadsheet and an MARR of 15% per year.
• Plot two graphs: PW versus i values for both alternatives’ cash flows and PW versus i
values for incremental cash flows.
• Estimate the breakeven ROR values from both graphs, and use this estimate to select one
alternative.
Air Cleanser
(Filter 1)
Purely Heaven
(Filter 2)
Initial cost per aircraft, $ 1000 1500
Estimated savings, $ per year 375 700 in year 1, decreasing by
100 per year thereafter
Estimated life, years 5 5
Solution by Spreadsheet
Refer to Figure 8–6 as the solution is explained. For information, row 10 shows the ROR values
calculated using the IRR function for filter 1 and filter 2 cash flows and the incremental cash
flow series (filter 2 filter 1).
The cash flow sign tests for each filter indicate no multiple rates. The incremental cash flow
sign test does not indicate the presence of a unique positive root; however, the second rate is an
extremely large and useless value. The PW values of filter 1 and filter 2 cash flows are plotted
on the right side for i values ranging from 0% to 60%. Since the PW curves cross each other at
approximately i * 17%, the rate does exceed the MARR of 15%. The higher-cost filter 2
(Purely Heaven) is selected.
The PW curve for the incremental cash flows series (column G) is plotted at the bottom left
of Figure 8–6 . As expected, the curve crosses the PW 0 line at approximately 17%, indicating
the same economic conclusion of filter 2.

MARR
Breakeven
Incremental ROR 17%
Figure 8–6
PW versus i graph and PW versus incremental i graph, Example 8.4 .
Breakeven ROR 17%
MARR
Filter 1 ROR 25%
Filter 2 ROR 23%
212

8.5 Rate of Return Evaluation Using AW 213
Figure 8–6 provides an excellent opportunity to see why the ROR method can result in selecting
the wrong alternative when only i * values are used to select between two alternatives. This is
sometimes called the ranking inconsistency problem of the ROR method. The inconsistency
occurs when the MARR is set less than the breakeven rate between two revenue alternatives.
Since the MARR is established based on conditions of the economy and market, MARR is established
external to any particular alternative evaluation. In Figure 8–6 the incremental breakeven
rate is 16.89%, and the MARR is 15%. The MARR is lower than breakeven; therefore, the incremental
ROR analysis results in correctly selecting filter 2. But if only the i * values were used,
filter 1 would be wrongly chosen, because its i * exceeds that of filter 2 (25.41% > 23.57%). This
error occurs because the rate of return method assumes reinvestment at the alternative’s ROR
value, while PW and AW analyses use the MARR as the reinvestment rate. The conclusion is
simple:
If the ROR method is used to evaluate two or more alternatives, use the incremental cash flows
and i* to make the decision between alternatives.
8.5 Rate of Return Evaluation Using AW
Comparing alternatives by the ROR method (correctly performed) always leads to the same selection
as PW and AW analyses, whether the ROR is determined using a PW-based or an AW-based
relation. However, for the AW-based technique, there are two equivalent ways to perform the
evaluation: (1) using the incremental cash flows over the LCM of alternative lives, just as for the
PW-based relation (Section 8.4), or (2) finding the AW for each alternative’s actual cash flows and
setting the difference of the two equal to zero to find the i * value. There is no difference between
the two approaches if the alternative lives are equal. Both methods are summarized here.
Since the ROR method requires comparison for equal service, the incremental cash flows
must be evaluated over the LCM of lives. There may be no real computational advantage to
using AW, as was found in Chapter 6. The same six-step procedure of the previous section (for
PW-based calculation) is used, except in step 5 the AW-based relation is developed.
Equal-service
requirement
The second AW-based method mentioned above takes advantage of the AW technique’s assum p-
tion that the equivalent AW value is the same for each year of the first and all succeeding life
cycles. Whether the lives are equal or unequal, set up the AW relation for the cash fl ows of each
alternative, form the relation below, and solve for i *.
0 AW B AW A [8.3]
For both methods, all equivalent values are on an AW basis, so the i * that results from Equation
[8.3] is the same as the i * found using the first approach. Example 8.5 illustrates ROR
analysis using AW-based relations for unequal lives.
EXAMPLE 8.5
Compare the alternatives of vendors A and B for Ford in Example 8.3 , using an AW-based
incremental ROR method and the same MARR of 12% per year.
Solution
For reference, the PW-based ROR relation, Equation [8.2], for the incremental cash flow in
Example 8.3 shows that vendor B should be selected with i * 12.65%.
For the AW relation, there are two equivalent solution approaches. Write an AW-based
relation on the incremental cash flow series over the LCM of 10 years, or write Equation [8.3]
for the two actual cash flow series over one life cycle of each alternative.
For the incremental method, the AW equation is
0 5000( A P , i *,10) 11,000( P F , i *,5)( A P , i *,10) 2000( A F , i *,10) 1900

214 Chapter 8 Rate of Return Analysis: Multiple Alternatives
It is easy to enter the incremental cash flows onto a spreadsheet, as in Figure 8–3 , column D,
and use the IRR(D4:D14) function to display i * 12.65%.
For the second method, the ROR is found using the actual cash flows and the respective
lives of 10 years for A and 5 years for B.
Now develop 0 AW B AW A .
AW A 8000( A P , i ,10) 3500
AW B 13,000( A P , i ,5) 2000( A F , i ,5) 1600
0 13,000( A P , i *,5) 2000( A F , i *,5) 8000( A P , i *,10) 1900
Solution again yields i * 12.65%.
Comment
It is very important to remember that when an incremental ROR analysis using an AW-based
equation is made on the incremental cash fl ows, the LCM must be used.
8.6 Incremental ROR Analysis of Multiple Alternatives
This section treats selection from multiple alternatives that are mutually exclusive, using the incremental
ROR method. Acceptance of one alternative automatically precludes acceptance of
any others. The analysis is based upon PW (or AW) relations for incremental cash flows between
two alternatives at a time.
When the incremental ROR method is applied, the entire investment must return at least the
MARR. When the i * values on several alternatives exceed the MARR, incremental ROR evaluation
is required. (For revenue alternatives, if not even one i * MARR, the do-nothing alternative is selected.)
For all alternatives (revenue or cost), the incremental investment must be separately justified.
If the return on the extra investment equals or exceeds the MARR, then the extra investment should
be made in order to maximize the total return on the money available, as discussed in Section 8.1.
For ROR analysis of multiple, mutually exclusive alternatives, the following criteria are used.
ME alternative
selection
Select the one alternative
That requires the largest investment, and
Indicates that the extra investment over another acceptable alternative is justified.
An important rule to apply when evaluating multiple alternatives by the incremental ROR method
is that an alternative should never be compared with one for which the incremental investment is
not justifi ed.
The incremental ROR evaluation procedure for multiple, equal-life alternatives is summarized
below. Step 2 applies only to revenue alternatives, because the first alternative is compared
to DN only when revenue cash flows are estimated. The terms defender and challenger are dynamic
in that they refer, respectively, to the alternative that is currently selected (the defender)
and the one that is challenging it for acceptance based on i *. In every pairwise evaluation, there
is one of each. The steps for solution by hand or by spreadsheet are as follows:
1. Order the alternatives from smallest to largest initial investment. Record the annual cash
flow estimates for each equal-life alternative.
2. Revenue alternatives only: Calculate i * for the first alternative. In effect, this makes DN the
defender and the first alternative the challenger. If i * MARR, eliminate the alternative and
go to the next one. Repeat this until i * MARR for the first time, and define that alternative
as the defender. The next alternative is now the challenger. Go to step 3. ( Note: This is where
solution by spreadsheet can be a quick assist. Calculate the i * for all alternatives first using
the IRR function, and select as the defender the first one for which i * MARR. Label it the
defender and go to step 3.)
3. Determine the incremental cash flow between the challenger and defender, using the relation
Incremental cash flow challenger cash flow defender cash flow
Set up the ROR relation.

8.6 Incremental ROR Analysis of Multiple Alternatives 215
4. Calculate i * for the incremental cash flow series using a PW- or AW-based equation. (PW
is most commonly used.)
5. If i * MARR, the challenger becomes the defender and the previous defender is eliminated.
Conversely, if i * MARR, the challenger is removed, and the defender remains
against the next challenger.
6. Repeat steps 3 to 5 until only one alternative remains. It is the selected one.
Note that only two alternatives are compared at any one time. It is vital that the correct alternatives
be compared, or the wrong alternative may be selected.
EXAMPLE 8.6
Caterpillar Corporation wants to build a spare parts storage facility in the Phoenix, Arizona,
vicinity. A plant engineer has identified four different location options. The initial cost of earthwork
and prefab building and the annual net cash flow estimates are detailed in Table 8–5. The
annual net cash flow series vary due to differences in maintenance, labor costs, transportation
charges, etc. If the MARR is 10%, use incremental ROR analysis to select the one economically
best location.
TABLE 8–5 Estimates for Four Alternative Building Locations, Example 8.6
A B C D
Initial cost, $ 200,000 275,000 190,000 350,000
Annual cash flow, $ per year 22,000 35,000 19,500 42,000
Life, years 30 30 30 30
Solution
All sites have a 30-year life, and they are revenue alternatives. The procedure outlined above
is applied.
1. The alternatives are ordered by increasing initial cost in Table 8–6.
2. Compare location C with the do-nothing alternative. The ROR relation includes only the
P A factor.
0 190,000 19,500( P A , i *,30)
Table 8–6, column 1, presents the calculated ( P A , i *,30) factor value of 9.7436 and
i c
* 9.63%. Since 9.63% 10%, location C is eliminated. Now the comparison is A to
DN, and column 2 shows that i A
* 10.49%. This eliminates the do-nothing alternative;
the defender is now A and the challenger is B.
TABLE 8–6 Computation of Incremental Rate of Return for Four Alternatives,
Example 8.6
C
(1)
Initial cost, $ 190,000 200,000 275,000 350,000
Cash flow, $ per year 19,500 22,000 35,000 42,000
Alternatives compared C to DN A to DN B to A D to B
Incremental cost, $ 190,000 200,000 75,000 75,000
Incremental cash flow, $ 19,500 22,000 13,000 7,000
Calculated ( P A , i *,30) 9.7436 9.0909 5.7692 10.7143
i *,% 9.63 10.49 17.28 8.55
Increment justified? No Yes Yes No
Alternative selected DN A B B
A
(2)
B
(3)
D
(4)

216 Chapter 8 Rate of Return Analysis: Multiple Alternatives
3. The incremental cash flow series, column 3, and i * for B-to-A comparison are determined
from
0 275,000 (200,000) (35,000 22,000)( P A , i *,30)
75,000 13,000( P A , i *,30)
4. From the interest tables, look up the P A factor at the MARR, which is ( P A ,10%,30)
9.4269. Now, any P A value greater than 9.4269 indicates that the i * will be less than
10% and is unacceptable. The P A factor is 5.7692, so B is acceptable. For reference
purposes, i * 17.28%.
5. Alternative B is justified incrementally (new defender), thereby eliminating A.
6. Comparison D-to-B (steps 3 and 4) results in the PW relation 0 75,000
7000( P A , i *,30) and a P A value of 10.7143 ( i * 8.55%). Location D is eliminated,
and only alternative B remains; it is selected.
Comment
An alternative must always be incrementally compared with an acceptable alternative, and the
do-nothing alternative can end up being the only acceptable one. Since C was not justified in
this example, location A was not compared with C. Thus, if the B-to-A comparison had not
indicated that B was incrementally justified, then the D-to-A comparison would be correct
instead of D-to-B.
To demonstrate how important it is to apply the ROR method correctly, consider the following.
If the i * of each alternative is computed initially, the results by ordered alternatives are
Location C A B D
i *, % 9.63 10.49 12.35 11.56
Now apply only the first criterion stated earlier; that is, make the largest investment that has a
MARR of 10% or more. Location D is selected. But, as shown above, this is the wrong selection,
because the extra investment of $75,000 over location B will not earn the MARR. In fact,
it will earn only 8.55%. This is another example of the ranking inconsistency problem of the
ROR method mentioned in Section 8.4.
For cost alternatives, the incremental cash flow is the difference between costs for two alternatives.
There is no do-nothing alternative and no step 2 in the solution procedure. Therefore, the
lowest-investment alternative is the initial defender against the next-lowest investment (c hallenger).
This procedure is illustrated in Example 8.7 using a spreadsheet solution.
EXAMPLE 8.7
The complete failure of an offshore platform and the resulting spillage of up to 800,000 to
1,000,000 gallons per day into the Gulf of Mexico in the spring of 2010 have made major oil producers
and transporters very conscious of the harm done to people’s livelihood and all forms of
aquatic life by spills of this magnitude. To address the specific danger to birds that are shoreline
feeders and dwellers, environmental engineers from several international petroleum corporations
and transport companies—Exxon-Mobil, BP, Shell, and some transporters for OPEC producers—
have developed a plan to strategically locate throughout the world newly developed equipment
that is substantially more effective than manual procedures in cleaning crude oil residue from bird
feathers. The Sierra Club, Greenpeace, and other international environmental interest groups are
in favor of the initiative. Alternative machines from manufacturers in Asia, America, Europe, and
Africa are available with the cost estimates in Table 8–7. Annual cost estimates are expected to be
high to ensure readiness at any time. The company representatives have agreed to use the average
of the corporate MARR values, which results in MARR 13.5%. Use a spreadsheet and incremental
ROR analysis to determine which manufacturer offers the best economic choice.

8.6 Incremental ROR Analysis of Multiple Alternatives 217
TABLE 8–7 Costs for Four Alternative Machines, Example 8.7
Machine 1 Machine 2 Machine 3 Machine 4
First cost, $ 5,000 6,500 10,000 15,000
Annual operating cost, $ 3,500 3,200 3,000 1,400
Salvage value, $ 500 900 700 1,000
Life, years 8 8 8 8
Solution by Spreadsheet
Follow the procedure for incremental ROR analysis. The spreadsheet in Figure 8–7 contains
the complete solution.
1. The alternatives are already ordered by increasing first costs.
2. These are cost alternatives, so there is no comparison to DN, since i * values cannot be
calculated.
3. Machine 2 is the first challenger to machine 1; the incremental cash flows for the 2-to-1
comparison are in column D.
4. The 2-to-1 comparison results in i * 14.57% by applying the IRR function.
5. This return exceeds MARR 13.5%, so machine 2 is the new defender (cell D17).
The comparison continues for 3-to-2 in column E, where the return is negative at i *
18.77%; machine 2 is retained as the defender. Finally the 4-to-2 comparison has an incremental
ROR of 13.60%, which is slightly larger than MARR 13.5%. The conclusion is to
purchase machine 4 because the extra investment is (marginally) justified.
IRR(D6:D14)
Figure 8–7
Spreadsheet solution to select from multiple cost alternatives, Example 8.7 .
Comment
As mentioned earlier, it is not possible to generate a PW versus i graph for each cost alternative
because all cash flows are negative. However, it is possible to generate PW versus i graphs
for the incremental series in the same fashion as we have done previously. The curves will
cross the PW 0 line at the i * values determined by the IRR functions.
Selection from multiple, mutually exclusive alternatives with unequal lives using i * values
requires that the incremental cash flows be evaluated over the LCM of the two alternatives being
compared. This is another application of the principle of equal-service comparison. The spreadsheet
application in the next section illustrates the computations.
It is always possible to rely on PW or AW analysis of the incremental cash flows at the MARR
to make the selection. In other words, don’t find i * for each pairwise comparison; find PW or

218 Chapter 8 Rate of Return Analysis: Multiple Alternatives
AW at the MARR instead. However, it is still necessary to make the comparison over the LCM
number of years for an incremental analysis to be performed correctly.
8.7 All-in-One Spreadsheet Analysis (Optional)
For professors and students who like to pack a spreadsheet, Example 8.8 combines many of the
economic analysis techniques we have learned so far—(internal) ROR analysis, incremental
ROR analysis, PW analysis, and AW analysis. Now that the IRR, NPV, and PV functions are
mastered, it is possible to perform a wide variety of evaluations for multiple alternatives on a
single spreadsheet. No cell tags are provided in this example. A nonconventional cash flow series
for which multiple ROR values may be found, and selection from both mutually exclusive alternatives
and independent projects, are included in this example.
EXAMPLE 8.8
In-flight texting, phone, and Internet connections provided at airline passenger seats are an
expected service by many customers. Singapore Airlines knows it will have to replace 15,000
to 24,000 units in the next few years on its Boeing 757, 777, and A300 aircraft. Four optional
data handling features that build upon one another are available from the manufacturer, but at
an added cost per unit. Besides costing more, the higher-end options (e.g., satellite-based plugin
video service) are estimated to have longer lives before the next replacement is forced by
new, advanced features expected by flyers. All four options are expected to boost annual revenues
by varying amounts. Figure 8–8 spreadsheet rows 2 through 6 include all the estimates
for the four options.
(a) Using MARR 15%, perform ROR, PW, and AW evaluations to select the one level of
options that is the most promising economically.
(b) If more than one level of options can be selected, consider the four that are described as
independent projects. If no budget limitations are considered at this time, which options
are acceptable if the MARR is increased to 20% when more than one option may be
implemented?
Figure 8–8
Spreadsheet analysis using ROR, PW, and AW methods for unequal-life, revenue alternatives, Example 8.8 .

Chapter Summary 219
Solution by Spreadsheet
(a) The spreadsheet ( Figure 8–8 ) is divided into six sections:
Section 1 (rows 1, 2): MARR value and the alternative names (A through D) are in increasing
order of initial cost.
Section 2 (rows 3 to 6): Per-unit net cash flow estimates for each alternative. These are
revenue alternatives with unequal lives.
Section 3 (rows 7 to 20): Actual and incremental cash flows are displayed here.
Section 4 (rows 21, 22): Because these are all revenue alternatives, i * values are determined
by the IRR function. If an alternative passes the MARR test ( i * 15%), it is
retained and a column is added to the right of its actual cash flows so the incremental
cash flows can be determined. Columns F and H were inserted to make space for the
incremental evaluations. Alternative A does not pass the i * test.
Section 5 (rows 23 to 25): The IRR functions display the i * values in columns F and H.
Comparison of C to B takes place over the LCM of 12 years. Since i * CB 19.42%
15%, eliminate B; alternative C is the new defender and D is the next challenger. The
final comparison of D to C over 12 years results in i * DC 11.23% 15%, so D is
eliminated. Alternative C is the chosen one.
Section 6 (rows 26 to 28): These include the AW and PW analyses. The AW value over the life
of each alternative is calculated using the PMT function at the MARR with an embedded
NPV function. Also, the PW value is determined from the AW value for 12 years using the
PV function. For both measures, alternative C has the numerically largest value, as expected.
Conclusion: All methods result in the same, correct choice of alternative C.
(b) Since each option is independent of the others, and there is no budget limitation at this
time, each i * value in row 21 of Figure 8–8 is compared to MARR 20%. This is a comparison
of each option with the do-nothing alternative. Of the four, options B and C have
i * 20%. They are acceptable; the other two are not.
Comment
In part ( a ), we should have applied the two multiple-root sign tests to the incremental cash flow
series for the C-to-B comparison. The series itself has three sign changes, and the cumulative
cash flow series starts negatively and also has three sign changes. Therefore, up to three realnumber
roots may exist. The IRR function is applied in cell F23 to obtain i * CB 19.42%
without using a supplemental (Section 7.5) procedure. This means that the investment assumption
of 19.42% for positive cash flows is a reasonable one. If the MARR 15%, or some other
earning rate were more appropriate, the ROIC procedure could be applied to determine a single
rate, which would be different from 19.42%. Depending upon the investment rate chosen,
alternative C may or may not be incrementally justified against B. Here, the assumption is
made that the i * value is reasonable, so C is justified.
CHAPTER SUMMARY
Just as present worth and annual worth methods find the best alternative from among several,
incremental rate of return calculations can be used for the same purpose. In using the ROR technique,
it is necessary to consider the incremental cash flows when selecting between mutually
exclusive alternatives. The incremental investment evaluation is conducted between only two
alternatives at a time, beginning with the lowest initial investment alternative. Once an alternative
has been eliminated, it is not considered further.
Rate of return values have a natural appeal to management, but the ROR analysis is often
more difficult to set up and complete than the PW or AW analysis using an established MARR.
Care must be taken to perform a ROR analysis correctly on the incremental cash flows; otherwise
it may give incorrect results.
If there is no budget limitation when independent projects are evaluated, the ROR value of
each project is compared to the MARR. Any number, or none, of the projects can be accepted.

220 Chapter 8 Rate of Return Analysis: Multiple Alternatives
PROBLEMS
Understanding Incremental ROR
8.1 In a tabulation of cash flow, the column entitled
“Rate of return on the incremental cash flow” represents
the rate of return on what?
8.2 If the rate of return on the incremental cash flow
between two alternatives is less than the minimum
attractive rate of return, which alternative should
be selected, if any?
8.3 An engineer is comparing three projects by the incremental
ROR method. There are revenue and
cost cash flow estimates. He used the IRR function
of Excel to determine the ROR values for
each project. The first one is 3.5% above the
MARR, the second is 1.2% below the MARR, and
the third is 2.4% above the MARR. Which alternatives,
if any, must he include in the incremental
ROR analysis?
8.4 The rates of return on alternatives X and Y are
15% and 12%, respectively. Alternative Y requires
a larger investment than alternative X.
(a)
What is known about the rate of return on the
increment of investment between the two alternatives?
(b) If the MARR is 12%, which alternative
should be selected and why?
8.5 Victoria is comparing two mutually exclusive
alternatives, A and B. The overall ROR on alternative
A is greater than the MARR, and the overall
ROR on alternative B, which requires the larger
investment, is exactly equal to the MARR.
(a) What is known about the ROR on the increment
between A and B?
(b) Which alternative should be selected?
8.6 A food processing company is considering two
types of moisture analyzers. Only one can be selected.
The company expects an infrared model to
yield a rate of return of 27% per year. A more expensive
microwave model will yield a rate of return
of 22% per year. If the company’s MARR is
19% per year, can you determine which model
should be purchased solely on the basis of the
overall rate of return information provided? Why
or why not?
8.7 If $80,000 is invested at 30% and another $50,000
is invested at 20% per year, what is the overall rate
of return on the entire $130,000?
8.8 A total of $100,000 was invested in two different
projects identified as Z1 and Z2. If the overall rate
of return on the $100,000 was 30% and the rate of
return on the $30,000 invested in Z1 was 15%,
what was the rate of return on Z2?
8.9 Tuggle, Inc., which manufactures rigid shaft couplings,
has $600,000 to invest. The company is
considering three different projects that will yield
the following rates of return.
Project X i X 24%
Project Y i Y 18%
Project Z i Z 30%
The initial investment required for each project is
$100,000, $300,000, and $200,000, respectively.
If Tuggle’s MARR is 15% per year and it invests
in all three projects, what rate of return will the
company make?
8.10 Two options are available for setting up a wireless
meter scanner and controller. A simple setup is
good for 2 years and has an initial cost of $12,000,
no salvage value, and an operating cost of $27,000
per year. A more permanent system has a higher
first cost of $73,000, but it has an estimated life of
6 years and a salvage value of $15,000. It costs
only $14,000 per year to operate and maintain. If
the two options are compared using an incremental
rate of return, what are the incremental cash flows
in ( a ) year 0 and ( b ) year 2?
8.11 Prepare a tabulation of incremental cash flows for
the two machine alternatives below.
Machine X
Machine Y
First cost, $ 35,000 90,000
Annual operating cost, $ per year 31,600 19,400
Salvage value, $ 0 8,000
Life, years 2 4
8.12 For the alternatives shown, determine the sum of
the incremental cash flows for Q – P.
Alternative P Alternative Q
First cost, $ 50,000 85,000
Annual operating cost, $ per year 8,600 2,000
Annual revenue, $ per year 22,000 45,000
Salvage value, $ 3,000 8,000
Life, years 3 6
8.13 The tabulation of the incremental cash flows between
alternatives A and B is shown on the next
page. Alternative A has a 3-year life and alternative
B a 6-year life. If neither alternative has a salvage
value, what is ( a ) the first cost of alternative
A and ( b ) the first cost of alternative B?

Problems 221
Year
Incremental
Cash Flow (B A), $
0 –20,000
1 5,000
2 5,000
3 12,000
4 5,000
5 5,000
6 5,000
8.14 Standby power for pumps at water distribution
booster stations can be provided by either gasolineor
diesel-powered engines. The costs for the gasoline
engines are as follows:
Gasoline
First cost, $ 150,000
Annual M&O, $ per year 41,000
Salvage value, $ 23,000
Life, years 15
The incremental PW cash flow equation associated
with (diesel – gasoline) is
0 40,000 11,000( P A , i ,15) 16,000( P F , i ,15)
Determine the following:
( a) First cost of the diesel engines
( b) Annual M&O cost of the diesel engines
( c) Salvage value of the diesel engines
8.15 Several high-value parts for NASA’s reusable
space exploration vehicle can be either anodized
or powder-coated. Some of the costs for each
process are shown below:
Anodize
Powder Coat
First cost, $ ? 65,000
Annual cost, $ per year 21,000 ?
Resale value, $ ? 6,000
Life, years 3 3
The incremental AW cash flow equation associated
with (powder coat – anodize) is
0 –14,000( A P , i ,3) 5000 2000( A F , i ,3)
What is ( a ) the first cost for anodizing, ( b ) the annual
cost for powder coating, and ( c ) the resale
(salvage) value of the anodized parts?
Incremental ROR Comparison (Two Alternatives)
8.16 Specialty Gases & Chemicals manufactures nitrogen
trifluoride, a highly specialized gas used as an
industrial cleansing agent for flat panels installed in
laptop computers, televisions, and desktop monitors.
The incremental cash flow associated with two alternatives
for chemical storage and handling systems
(identified as P3 and X3) has been calculated in
$1000 units. Determine ( a ) the rate of return on the
incremental cash flows and ( b ) which one should be
selected if the company’s MARR is 25% per year.
Alternative X3 requires the larger initial investment.
Year
Incremental Cash Flow
(X3 P3), $1000
0 $4600
1–9 1100
10 2000
8.17 As groundwater wells age, they sometimes begin to
pump sand (and they become known as “sanders”),
and this can cause damage to downstream desalting
equipment. This situation can be dealt with by drilling
a new well at a cost of $1,000,000 or by installing
a tank and self-cleaning screen ahead of the
desalting equipment. The tank and screen will cost
$230,000 to install and $61,000 per year to operate
and maintain. A new well will have a pump that is
more efficient than the old one, and it will require
almost no maintenance, so its operating cost will be
only $18,000 per year. If the salvage values are estimated
at 10% of the first cost, use a present worth
relation to ( a ) calculate the incremental rate of return
and ( b ) determine which alternative is better at
a MARR of 6% per year over a 20-year study
period.
8.18 Konica Minolta plans to sell a copier that prints
documents on both sides simultaneously, cutting in
half the time it takes to complete big commercial jobs.
The costs associated with producing chemicallytreated
vinyl rollers and fiber-impregnated rubber
rollers are shown below. Determine which of
the two types should be selected by calculating the
rate of return on the incremental investment.
Assume the company’s MARR is 21% per year.
(Solve by hand and/or spreadsheet, as instructed.)
Treated
Impregnated
First cost, $ 50,000 95,000
Annual cost, $ per year 100,000 85,000
Salvage value, $ 5,000 11,000
Life, years 3 6
8.19 The Texas Department of Transportation (TxDOT)
is considering two designs for crash barriers along
a reconstructed portion of I-10. Design 2B will
cost $3 million to install and $135,000 per year to
maintain. Design 4R will cost $3.7 million to install
and $70,000 per year to maintain. Calculate
the rate of return and determine which design is
preferred if TxDOT uses a MARR of 6% per year
and a 20-year project period.

222 Chapter 8 Rate of Return Analysis: Multiple Alternatives
8.20 Chem-Tex Chemical is considering two additives
for improving the dry-weather stability of its lowcost
acrylic paint. Additive A has a first cost of
$110,000 and an annual operating cost of $60,000.
Additive B has a first cost of $175,000 and an annual
operating cost of $35,000. If the company
uses a 3-year recovery period for paint products
and a MARR of 20% per year, which process is
economically favored? Use an incremental ROR
analysis.
8.21 The manager of Liquid Sleeve, Inc., a company
that makes a sealing solution for machine shaft surfaces
that have been compromised by abrasion,
high pressures, or inadequate lubrication, is considering
adding metal-based nanoparticles of either
type Al or Fe to its solution to increase the product’s
performance at high temperatures. The costs
associated with each are shown below. The company’s
MARR is 20% per year. Do the following
using a PW-based rate of return analysis and a
spreadsheet:
(a)
Determine which nanoparticle type the company
should select using the i * value.
(b) On the same graph, plot the PW versus different
i values for each alternative. Indicate
the breakeven i * value and the MARR value
on the plot.
(c)
Use the plot of PW versus i values to select
the better alternative with MARR 20% per
year. Is the answer the same as in part ( a )?
Type Fe
Type Al
First cost, $ 150,000 280,000
Annual operating cost, $ per year 92,000 74,000
Salvage value, $ 30,000 70,000
Life, years 2 4
8.22 A chemical company is considering two processes
for isolating DNA material. The incremental cash
flows between the two alternatives, J and S, have an
incremental rate of return that is less than 40%,
which is the MARR of the company. However, the
company CEO prefers the more expensive process
S. She believes the company can implement cost
controls to reduce the annual cost of the more
expensive process. By how much would she have to
reduce the annual operating cost of alternative S (in
$ per year) for it to have an incremental rate of
return of exactly 40%?
Year
Incremental Cash Flow
(S J), $
0 900,000
1 400,000
2 400,000
3 400,000
8.23 The incremental cash flows for two alternative
electrode setups are shown. The MARR is 12%
per year, and alternative Dryloc requires a larger
initial investment compared to NPT.
(a)
Determine which should be selected using an
AW-based rate of return analysis.
(b) Use a graph of incremental values to determine
the largest MARR value that will justify
the NPT alternative.
Year
Incremental Cash Flow
(Dryloc NPT), $
0 56,000
18 8,900
9 12,000
8.24 Hewett Electronics manufactures amplified pressure
transducers. It must decide between two machines
for a finishing operation. Select one for them
on the basis of AW-based rate of return analysis.
The company’s MARR is 18% per year.
Variable
Speed
Dual
Speed
First cost, $ 270,000 245,000
Annual operating cost, $ per year 135,000 139,000
Salvage value, $ 75,000 35,000
Life, years 6 6
8.25 A manufacturer of hydraulic equipment is trying to
determine whether it should use monoflange double
block and bleed (DBB) valves or a multi-valve
system (MVS) for chemical injection. The costs
are shown below. Use an AW-based rate of return
analysis and a MARR of 18% per year to determine
the better of the two options.
DBB
MVS
First cost, $ 40,000 71,000
Annual cost, $ per year 60,000 65,000
Salvage value, $ 0 18,000
Life, years 2 4
8.26 Poly-Chem Plastics is considering two types of injection
molding machines—hydraulic and electric.
The hydraulic press (HP) will have a first cost of
$600,000, annual costs of $200,000, and a salvage
value of $70,000 after 5 years. Electric machine
technology (EMT) will have a first cost of $800,000,
annual costs of $150,000, and a salvage value of
$130,000 after 5 years.
(a) Use an AW-based rate of return equation to
determine the ROR on the increment of
investment between the two.

Problems 223
(b) Determine which machine the company
should select, if the MARR 16% per year.
( c) Plot the AW versus i graph for each alternative’s
cash flows, and utilize it to determine
the largest MARR that will justify the EMT
extra investment of $200,000.
8.27 Last week Eduardo calculated the overall project
ROR values for two alternatives A and B using the
estimates below. He calculated i A
* 34.2% and
i B
* 31.2% and recommended acceptance of A
since its rate of return exceeded the established
MARR of 30% by a greater amount than project B.
Yesterday, the general manager of the company
announced a major capital investment program,
which includes a large drop in the MARR from
30% to 20% per year. Do the following to help
Eduardo better understand the rate of return method
and what this reduction in MARR means.
( a) Explain the error that Eduardo made in performing
the rate of return analysis.
( b) Perform the correct analysis using each
MARR value.
( c) Illustrate the ranking inconsistency problem
using the two MARR values, and determine
the maximum MARR that will justify alternative
B.
Alternative A
Alternative B
First cost, $ 40,000 85,000
Annual operating cost, 5,500 15,000
$ per year
Annual revenue, $ per year 22,000 45,000
Salvage value, $ 0 20,000
Life, years 6 6
i *, % 34.2 29.2
Multiple Alternative Comparison
8.28 Four mutually exclusive revenue alternatives
are under consideration to automate a baking
and packaging process at Able Bakery Products.
The alternatives are ranked in order of increasing
initial investment and compared by
incremental rate of return analysis. The rate of
return on each increment of investment was less
than the MARR. Which alternative should be
selected?
8.29 A WiMAX wireless network integrated with a
satellite network can provide connectivity to
any location within 10 km of the base station.
The number of sectors per base station can be
varied to increase the bandwidth. An independent
cable operator is considering three bandwidth
alternatives. Assume a life of 20 years and
a MARR of 10% per year to determine which
alternative is best using an incremental rate of
return analysis.
Bandwidth,
Mbps
First Cost,
$1000
Operating
Cost,
$1000 per Year
Annual
Income,
$1000 per Year
44 –40,000 –2000 5000
55 –46,000 –1000 5000
88 –61,000 –500 8000
8.30 Xerox’s iGenX high-speed commercial printers
cost $1.5 billion to develop. The machines cost
$500,000 to $750,000 depending on what options
the client selects. Spectrum Imaging Systems is
considering the purchase of a new printer based
on recent contracts it received for printing weekly
magazine and mail-out advertising materials. The
operating costs and revenues generated are related
to a large extent to the speed and other capabilities
of the copier. Spectrum is considering the four
machines shown below. The company uses a
3-year planning period and a MARR of 15% per
year. Determine which copier the company should
acquire on the basis of an incremental rate of return
analysis.
Copier
Initial
Investment,
$
Operating
Cost,
$ per Year
Annual
Revenue,
$ per Year
Salvage
Value, $
iGen-1 –500,000 –350,000 450,000 70,000
iGen-2 –600,000 –300,000 460,000 85,000
iGen-3 –650,000 –275,000 480,000 95,000
iGen-4 –750,000 –200,000 510,000 120,000
8.31 Ashley Foods, Inc. has determined that any one of
five machines can be used in one phase of its chili
canning operation. The costs of the machines are
estimated below, and all machines are estimated to
have a 4-year useful life. If the minimum attractive
rate of return is 20% per year, determine which
machine should be selected on the basis of a rate of
return analysis.
Machine First Cost, $
Annual
Operating
Cost, $ per Year
1 31,000 16,000
2 29,000 19,300
3 34,500 17,000
4 49,000 12,200
5 41,000 15,500
8.32 Five revenue projects are under consideration by
General Dynamics for improving material flow
through an assembly line. The initial cost in

224 Chapter 8 Rate of Return Analysis: Multiple Alternatives
$1000 and the life of each project are as follows
(revenue estimates are not shown):
Project
A B C D E
Initial cost, $1000 700 2300 900 300 1600
Life, years 5 8 5 5 6
An engineer made the comparisons shown below.
From the calculations, determine which project, if
any, should be undertaken if the company’s MARR
is ( a ) 11.5% per year and ( b ) 13.5% per year. If
other calculations are necessary to make a decision,
state which ones.
Comparison Incremental Rate of Return, %
B vs DN 13%
A vs B 19%
D vs DN 11%
E vs B 15%
E vs D 24%
E vs A 21%
C vs DN 7%
C vs A 19%
E vs DN 12%
A vs DN 10%
E vs C 33%
D vs C 33%
D vs B 29%
8.33 The five alternatives shown here are being evaluated
by the rate of return method.
Alternative
Initial
Investment,
$
ROR
versus
DN, %
Incremental
Rate of Return, %
A B C D E
A –25,000 9.6 — 27.3 9.4 35.3 25.0
B –35,000 15.1 — 0 38.5 24.4
C –40,000 13.4 — 46.5 27.3
D –60,000 25.4 — 6.8
E –75,000 20.2 —
(a)
(b)
If the alternatives are mutually exclusive and
the MARR is 26% per year, which alternative
should be selected?
If the alternatives are mutually exclusive and
the MARR is 15% per year, which alternative
should be selected?
(c) If the alternatives are independent and the
MARR is 15% per year, which alternative(s)
should be selected?
8.34 Five mutually exclusive revenue alternatives that
have infi nite live s are under consideration for increasing
productivity in a manufacturing operation.
The initial costs and cash flows of each
project are shown. If the MARR is 14.9% per year,
which alternative should be selected?
Alternative
A B C D E
Initial cost, $ 7,000 23,000 9,000 3,000 16,000
Cash flow, $ per 1,000 3,500 1,400 500 2,200
year
Rate of return
(vs DN), %
14.3 15.2 15.6 16.7 13.8
8.35 The plant manager at Automaton Robotics is looking
at the summarized incremental rate of return
information shown below for five mutually exclusive
alternatives, one of which must be chosen.
The table includes the overall ROR and the incremental
comparison of alternatives. Which alternative
is best if the minimum attractive rate of return
is ( a ) 15% per year and ( b ) 12% per year?
Alternative
First
Cost, $
Overall
ROR, %
Incremental Rate of Return, %
A B C D E
A 80,000 14 — 12 11 17 24
B 60,000 16 12 — 14 23 21
C 40,000 17 11 14 — 35 29
D 30,000 12 17 23 35 — 17
E 20,000 8 24 21 29 17 —
8.36 Only one of four different machines can be purchased
for Glass Act Products. An engineer performed
the following analysis to select the best
machine, all of which have a 10-year life. Which
machine, if any, should the company select at a
MARR of 22% per year?
Machine
1 2 3 4
Initial cost, $ 44,000 60,000 72,000 98,000
Annual cost, $ per year 70,000 64,000 61,000 58,000
Annual savings, 80,000 80,000 80,000 82,000
$ per year
Overall ROR, % 18.6 23.4 23.1 20.8
Machines compared 2 to 1 3 to 2 4 to 3
Incremental
16,000 12,000 26,000
investment, $
Incremental cash flow,
6,000 3,000 5,000
$ per year
ROR on increment, % 35.7 21.4 14.1
8.37 The U.S. Bureau of Reclamation is considering five
national park projects shown below, all of which can
be considered to last indefinitely. At a MARR of
7.5% per year, determine which should be selected, if
they are ( a ) independent and ( b ) mutually exclusive.
Project ID
First
Cost, $1000
Annual
Income, $1000
Rate of
Return, %
A 20,000 2000 10.0
B 10,000 1300 13.0
C 15,000 1000 6.6
D 70,000 4000 5.7
E 50,000 2600 5.2

Additional Problems and FE Exam Review Questions 225
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
8.38 When conducting a rate of return (ROR) analysis
involving multiple mutually exclusive alternatives,
the first step is to:
( a) Rank the alternatives according to decreasing
initial investment cost
( b) Rank the alternatives according to increasing
initial investment cost
( c) Calculate the present worth of each alternative
using the MARR
( d) Find the LCM between all of the alternatives
8.39 In comparing mutually exclusive alternatives by
the ROR method, you should:
( a) Find the ROR of each alternative and pick
the one with the highest ROR
( b) Select the alternative whose incremental
ROR is the highest
( c) Select the alternative with ROR MARR
that has the lowest initial investment cost
(d) Select the alternative with the largest initial investment
that has been incrementally justified
8.40 When comparing independent projects by the
ROR method, you should:
( a) Find the ROR of each project and pick the
ones with the highest ROR
( b) Select all projects that have an overall ROR
MARR
( c) Select the project with an overall ROR
MARR that involves the lowest initial
investment cost
(d) Select the project with the largest initial investment
that has been incrementally justified
8.41 Of the following scenarios, alternative Y requires a
higher initial investment than alternative X, and
the MARR is 20% per year. The only scenario that
requires an incremental investment analysis to
s elect an alternative is that:
( a) X has an overall ROR of 22% per year, and Y
has an overall ROR of 24% per year
( b) X has an overall ROR of 19% per year, and Y
has an overall ROR of 23% per year
( c) X has an overall ROR of 18% per year, and Y
has an overall ROR of 19% per year
( d) X has an overall ROR of 28% per year, and Y
has an overall ROR of 26% per year
8.42 Alternatives whose cash flows (excluding the
salvage value) are all negative are called:
( a) Revenue alternatives
( b) Nonconventional alternatives
( c) Cost alternatives
( d) Independent alternatives
8.43 For these alternatives, the sum of the incremental
cash flows is:
Year A B
0 10,000 14,000
1 2,500 4,000
2 2,500 4,000
3 2,500 4,000
4 2,500 4,000
5 2,500 4,000
(a) $2500
(b) $3500
(c) $6000
(d) $8000
8.44 Helical Systems, Inc. uses a minimum attractive
rate of return of 8% per year, compounded annually.
The company is evaluating two new processes
for improving the efficiency of its manufacturing
operation. The cash flow estimates
associated with each process are shown below. A
correct equation to use for an incremental rate of
return analysis is:
Process I
Process J
First cost, $ 40,000 50,000
Annual cost, $ per year 15,000 12,000
Salvage value, $ 5,000 6,000
Life, years 3 3
(a)
(b)
(c)
(d)
0 10,000 3000( P A , i *,3)
1000( P F , i *,3)
0 40,000( A P , i *,3) 15,000
5000( A F , i *,3)
0 50,000( A P , i *,3) 12,000
6000( A F , i *,3)
0 10,000 3000( P A , i *,3)
1000( P F , i *,3)
8.45 For the four independent projects shown, the one or
ones to select using a MARR of 14% per year are:
Project
Rate of Return,
% per Year
A 14
B 12
C 15
D 10
(a) Only C
(b) Only A and C
(c) Only A
( d) Can’t tell; need to conduct incremental
analysis

226 Chapter 8 Rate of Return Analysis: Multiple Alternatives
Problems 8.46 through 8.48 are based on the following
information.
Five alternatives are being evaluated by the incremental
rate of return method.
Alternative
Initial
Investment, $
Overall ROR
versus DN, %
Incremental
Rate of Return, %
A B C D E
A 25,000 9.6 — 27.3 9.4 35.3 25.0
B 35,000 15.1 — 0 38.5 24.4
C 40,000 13.4 — 46.5 27.3
D 60,000 25.4 — 6.8
E 75,000 20.2 —
8.46 If the projects are mutually exclusive and the minimum
attractive rate of return is 14% per year, the
best alternative is:
(a) B (b) C (c) D (d) E
8.47 If the projects are mutually exclusive and the
MARR is 20% per year, the best alternative is:
(a) B
( b) C
( c) D
( d) E
8.48 If the projects are independent, instead of mutually
exclusive, the one or ones to select at an MARR of
18% per year are:
( a) B and C
(b) B, D, and E
(c) D and E
(d) B, C, and E
CASE STUDY
ROR ANALYSIS WITH ESTIMATED LIVES THAT VARY
Background
Make-to-Specs is a software system under development by
ABC Corporation. It will be able to translate digital versions of
three-dimensional computer models, containing a wide variety
of part shapes with machined and highly finished (ultrasmooth)
surfaces. The product of the system is the numerically
controlled (NC) machine code for the part’s manufacturing.
Additionally, Make-to-Specs will build the code for superfine
finishing of surfaces with continuous control of the finishing
machines.
Information
There are two alternative computers that can provide the
server function for the software interfaces and shared database
updates on the manufacturing floor while Make-to-
Specs is operating in parallel mode. The server first cost
and estimated contribution to annual net cash flow are
summarized below.
Server 1 Server 2
First cost, $ 100,000 200,000
Net cash flow, $/year 35,000 50,000 year 1, plus 5000 per
year for years 2, 3, and 4
(gradient)
70,000 maximum for years
5 on, even if the server is
replaced
Life, years 3 or 4 5 or 8
The life estimates were developed by two different individuals:
a design engineer and a manufacturing manager. They have
asked that, at this stage of the project, all analyses be performed
using both life estimates for each system.
Case Study Exercises
Use spreadsheet analysis to determine the following:
1. If the MARR 12%, which server should be selected?
Use the PW or AW method to make the selection.
2. Use incremental ROR analysis to decide between the
servers at MARR 12%.
3. Use any method of economic analysis to display on the
spreadsheet the value of the incremental ROR between
server 2 with a life estimate of 5 years and a life estimate
of 8 years.

Case Study 227
CASE STUDY
HOW A NEW ENGINEERING GRADUATE CAN HELP HIS FATHER 1
Background
“I don’t know whether to sell it, expand it, lease it, or what. But
I don’t think we can keep doing the same thing for many more
years. What I really want to do is to keep it for 5 more years,
then sell it for a bundle,” Elmer Kettler said to his wife, Janise,
their son, John Kettler, and new daughter-in-law, Suzanne
Gestory, as they were gathered around the dinner table. Elmer
was sharing thoughts on Gulf Coast Wholesale Auto Parts, a
company he has owned and operated for 25 years on the southern
outskirts of Houston, Texas. The business has excellent contracts
for parts supply with several national retailers operating
in the area—NAPA, AutoZone, O’Reilly, and Advance. Additionally,
Gulf Coast operates a rebuild shop serving these same
retailers for major automobile components, such as carburetors,
transmissions, and air conditioning compressors.
At his home after dinner, John decided to help his father
with an important and difficult decision: What to do with his
business? John graduated just last year with an engineering
degree from a major state university in Texas, where he completed
a course in engineering economy. Part of his job at
Energcon Industries is to perform basic rate of return and
present worth analyses on energy management proposals.
Information
Over the next few weeks, John outlined five options, including
his dad’s favorite of selling in 5 years. John summarized
all the estimates over a 10-year horizon. The options and
estimates were given to Elmer, and he agreed with them.
Option 1: Remove rebuild. Stop operating the rebuild
shop and concentrate on selling wholesale
parts. The removal of the rebuild operations and the
switch to an “all-parts house” are expected to cost
$750,000 in the first year. Overall revenues will
drop to $1 million the first year with an expected 4%
increase per year thereafter. Expenses are projected
at $0.8 million the first year, increasing 6% per year
thereafter.
Option 2: Contract rebuild operations. To get the
rebuild shop ready for an operations contractor to
take over will cost $400,000 immediately. If expenses
stay the same for 5 years, they will average
$1.4 million per year, but they can be expected to
rise to $2 million per year in year 6 and thereafter.
Elmer thinks revenues under a contract arrangement
can be $1.4 million the first year and can rise 5% per
year for the duration of a 10-year contract.
Option 3: Maintain status quo and sell out after 5 years
(Elmer’s personal favorite). There is no cost now, but
the current trend of negative net profit will probably
continue. Projections are $1.25 million per year for expenses
and $1.15 million per year in revenue. Elmer
had an appraisal last year, and the report indicated Gulf
Coast Wholesale Auto Parts is worth a net $2 million.
Elmer’s wish is to sell out completely after 5 more
years at this price, and to make a deal that the new
owner pay $500,000 per year at the end of year 5 (sale
time) and the same amount for the next 3 years.
Option 4: Trade-out. Elmer has a close friend in the
antique auto parts business who is making a “ killing,”
so he says, with e-commerce. Although the possibility
is risky, it is enticing to Elmer to consider a whole new
line of parts, but still in the basic business that he already
understands. The trade-out would cost an estimated
$1 million for Elmer immediately. The 10-year
horizon of annual expenses and revenues is considerably
higher than for his current business. Expenses are
estimated at $3 million per year and revenues at
$3.5 million each year.
Option 5: Lease arrangement. Gulf Coast could be
leased to some turnkey company with Elmer remaining
the owner and bearing part of the expenses for building,
delivery trucks, insurance, etc. The first-cut estimates
for this option are $1.5 million to get the business ready
now, with annual expenses at $500,000 per year and
revenues at $1 million per year for a 10-year contract.
Case Study Exercises
Help John with the analysis by doing the following:
1. Develop the actual cash flow series and incremental
cash flow series (in $1000 units) for all five options in
preparation for an incremental ROR analysis.
2. Discuss the possibility of multiple rate of return values
for all the actual and incremental cash flow series. Find
any multiple rates in the range of 0% to 100%.
3. If John’s father insists that he make 25% per year or more
on the selected option over the next 10 years, what should
he do? Use all the methods of economic analysis you
have learned so far (PW, AW, ROR) so John’s father can
understand the recommendation in one way or another.
4. Prepare plots of the PW versus i for each of the five options.
Estimate the breakeven rate of return between options.
5. What is the minimum amount that must be received in
each of years 5 through 8 for option 3 (the one Elmer
wants) to be best economically? Given this amount,
what does the sale price have to be, assuming the same
payment arrangement as presented above?
1 Based upon a study by Alan C. Stewart, Consultant, Communications and High Tech Solutions Engineering, Accenture LLP.

CHAPTER 9
Benefit/Cost
Analysis and
Public Sector
Economics
LEARNING OUTCOMES
Purpose: Understand public sector projects and select the best alternative on the basis of incremental benefit /cost analysis.
SECTION TOPIC LEARNING OUTCOME
9.1 Public sector • Explain some of the fundamental differences
between private and public sector projects.
9.2 B/C for single project • Calculate the benefit/cost ratio and use it to
evaluate a single project.
9.3 Incremental B/C • Select the better of two alternatives using the
incremental B/C ratio method.
9.4 More than two alternatives • Based on the incremental B/C ratios, select the
best of multiple alternatives.
9.5 Service projects and CEA • Explain service sector projects and use costeffectiveness
analysis (CEA) to evaluate projects.
9.6 Ethical considerations • Explain the major aspects of public project
activities, and describe how ethical compromise
may enter public sector project analysis.

T
he evaluation methods of previous chapters are usually applied to alternatives
in the private sector, that is, for-profit and not-for-profit corporations and businesses.
This chapter introduces public sector and service sector alternatives and
their economic consideration. In the case of public projects, the owners and users (beneficiaries)
are the citizens and residents of a government unit—city, county, state, province,
or nation. Government units provide the mechanisms to raise capital and operating funds.
Public-private partnerships have become increasingly common, especially for large infrastructure
projects such as major highways, power generation plants, water resource developments,
and the like.
The benefit/cost (B/C) ratio introduces objectivity into the economic analysis of public sector
evaluation, thus reducing the effects of politics and special interests. The different formats
of B/C analysis, and associated disbenefits of an alternative, are discussed here. The B/C
analysis can use equivalency computations based on PW, AW, or FW values. Performed correctly,
the benefit/cost method will always select the same alternative as PW, AW, and ROR
analyses.
This chapter also introduces service sector projects and discusses how their economic evaluation
is different from that for other projects. Finally, there is a discussion on professional
ethics and ethical dilemmas in the public sector.
Water Treatment Facility #3 Case: Allen
Water Utilities has planned for the last
25 years to construct a new drinking
water treatment facility that will supply
the rapidly growing north and northwest
areas of the city. An expectation of
over 100,000 new residents in the next
several years and 500,000 by 2040
prompted the development of the plant
starting in 2012. The supply is from a
large surface lake currently used to provide
water to all of Allen and the surrounding
communities. The project is
termed WTF3, and its initial capital investment
is $540 million for the treatment
plant and two large steel-pipe
transmission mains (84- and 48-inch)
that will be installed via tunneling approximately
100 to 120 feet under suburban
areas of the city to reach current
r eservoirs.
Tunneling was selected after geotechnical
borings indicated that open trenching
was not supportable by the soil and
based upon a large public outcry against
trenching in the living areas along the
selected transmission routes. Besides the
treatment plant construction on the 95-
acre site, there must be at least three
large vertical shafts (25 to 50 feet in diameter)
bored along each transmission
main to gain underground access for
equipment and debris removal during
the tunneling operations.
PE
The stated criteria used to make decisions
for WTF3 and the transmission
mains were economics, environment,
community impact, and constructability.
There are major long-term benefits for
the new facility. These are some mentioned
by city engineers:
• It will meet projected water needs
of the city for the next 50 years.
• The new treatment plant is at a
higher elevation than the current
two plants, allowing gravity flow to
replenish reservoirs, thereby using
little or no electric pumping.
• There will be an increase in the
diversity and reliability of supply as
other plants age.
• It will provide a water quality that
is more consistent due to the location
of the raw water intakes.
• The facility uses water supplies
already purchased; therefore, there
is no need to negotiate additional
allowances.
The disbenefits are mostly short-term
during the construction of WTF3 and
transmission mains. Some of these are
mentioned by citizen groups and one retired
city engineer:
• There will be disruption of habitat
for some endangered species
of birds, lizards, and trees not

230 Chapter 9 Benefit/Cost Analysis and Public Sector Economics
found in any other parts of the
country.
• Large amounts of dust and smoke
will enter the atmosphere in a
residential area during the 3½ years
of construction, tunneling, and
transmission main completion.
• Noise pollution and traffic congestion
will result during an estimated
26,000 truck trips to remove debris
from the plant site and tunnel
shafts, in addition to the problems
from regular construction traffic.
• Natural landscape in plant and tunnel
shaft sites will be destroyed.
• Safety will be compromised for children
in a school where large trucks
will pass about every 5 minutes for
approximately 12 hours per day,
6 days per week for 2½ years.
• There may be delays in fire and
ambulance services in emergencies,
since many neighborhood streets are
country-road width and offer only
single ingress/egress streets for neighborhoods
along the indicated routes.
• The need for the facility has not
been proved, as the water will be
sold to developers outside the city
limits, not provided to residences
within Allen.
• Newly generated revenues will be
used to pay off the capital funding
bonds approved for the plant’s construction.
Last year, the city engineers did a benefit/cost
analysis for this massive public
sector project; none of the results were
publicized. Public and elected official intervention
has now caused some of the
conclusions using the criteria mentioned
above to be questioned by the general
manager of Allen Water Utilities.
This case is used in the following topics
of this chapter:
Public sector projects (Section 9.1)
Incremental B/C analysis, two alternatives
(Section 9.3)
Incremental B/C analysis, more than
two alternatives (Section 9.4)
9.1 Public Sector Projects
Virtually all the examples and problems of previous chapters have involved the private sector,
where products, systems, and services are developed and offered by corporations and businesses
for use by individual customers and clients, the government, or other companies. (Notable exceptions
are the long-life alternatives discussed in Chapters 5 (PW) and 6 (AW) where capitalized
cost analysis was applied.) Now we will explore projects that concentrate on government units
and the citizens they serve. These are called public sector projects.
A public sector project is a product, service, or system used, financed, and owned by the citizens
of any government level. The primary purpose is to provide service to the citizenry for the
public good at no profit. Areas such as public health, criminal justice, safety, transportation,
welfare, and utilities are publically owned and require economic evaluation.
Upon reflection, it is surprising how much of what we use on a daily or as-needed basis is
publicly owned and financed to serve us—the citizenry. These are some public sector
examples:
Hospitals and clinics
Parks and recreation
Utilities: water, electricity, gas,
sewer, sanitation
Schools: primary, secondary, community
colleges, universities
Economic development projects
Convention centers
Sports arenas
Transportation: highways, bridges,
waterways

9.1 Public Sector Projects 231
Police and fire protection
Courts and prisons
Food stamp and rent relief programs
Job training
Public housing
Emergency relief
Codes and standards
There are significant differences in the characteristics of private and public sector alternatives.
They are summarized here.
Characteristic Public sector Private sector
Size of investment Large Some large; more medium to small
Often alternatives developed to serve public needs require large initial investments, possibly
distributed over several years. Modern highways, public transportation systems, universities,
airports, and flood control systems are examples.
Characteristic Public sector Private sector
Life estimates Longer (30–50 years) Shorter (2–25 years)
The long lives of public projects often prompt the use of the capitalized cost method, where
infinity is used for n and annual costs are calculated as A P ( i ). As n gets larger, especially
over 30 years, the differences in calculated A values become small. For example, at i 7%,
there will be a very small difference in 30 and 50 years, because ( A/P ,7%,30) 0.08059 and
( A/P ,7%,50) 0.07246.
Characteristic Public sector Private sector
Annual cash flow
estimates
No profit; costs, benefits, and
disbenefits are estimated
Revenues contribute to
profits; costs are estimated
Public sector projects (also called publicly owned) do not have profits; they do have costs that are
paid by the appropriate government unit; and they benefit the citizenry. Public sector projects
often have undesirable consequences, as interpreted by some sectors of the public. It is these
consequences that can cause public controversy about the projects. The economic analysis should
consider these consequences in monetary terms to the degree estimable. (Often in private sector
analysis, undesirable consequences are not considered, or they may be directly addressed as
costs.) To perform a benefit/cost economic analysis of public alternatives, the costs (initial and
annual), the benefits, and the disbenefits, if considered, must be estimated as accurately as
possible in monetary units.
Costs—estimated expenditures to the government entity for construction, operation, and maintenance
of the project, less any expected salvage value.
Benefits—advantages to be experienced by the owners, the public.
Disbenefits—expected undesirable or negative consequences to the owners if the alternative is
implemented. Disbenefits may be indirect economic disadvantages of the alternative.
It is difficult to estimate and agree upon the economic impact of benefits and disbenefits for a
public sector alternative. For example, assume a short bypass around a congested area in town is
recommended. How much will it benefit a driver in dollars per driving minute to be able to bypass
five traffic lights while averaging 35 miles per hour, as compared to currently driving
through the lights averaging 20 miles per hour and stopping at an average of two lights for an
average of 45 seconds each? The bases and standards for benefits estimation are always difficult
to establish and verify. Relative to revenue cash flow estimates in the private sector, benefit estimates
are much harder to make, and vary more widely around uncertain averages. (The inability
to make economic estimates for benefits may be overcome by using the evaluation technique
discussed in Section 9.5.) And the disbenefits that accrue from an alternative are even harder to
estimate. In fact, the disbenefit itself may not be known at the time the evaluation is performed.

232 Chapter 9 Benefit/Cost Analysis and Public Sector Economics
Characteristic Public sector Private sector
Funding
Taxes, fees, bonds,
private funds
Stocks, bonds, loans,
individual owners
The capital used to finance public sector projects is commonly acquired from taxes, bonds,
and fees. Taxes are collected from those who are the owners—the citizens (e.g., federal gasoline
taxes for highways are paid by all gasoline users, and health care costs are covered by insurance
premiums). This is also the case for fees, such as toll road fees for drivers. Bonds are often issued:
U.S. Treasury bonds, municipal bond issues, and special-purpose bonds, such as utility district
bonds. Private lenders can provide up-front financing. Also, private donors may provide funding
for museums, memorials, parks, and garden areas through gifts.
Characteristic Public sector Private sector
Interest rate Lower Higher, based on cost of capital
Because many of the financing methods for public sector projects are classified as low-interest,
the interest rate is virtually always lower than for private sector alternatives. Government
agencies are exempt from taxes levied by higher-level units. For example, municipal projects
do not have to pay state taxes. (Private corporations and individual citizens do pay taxes.)
Many loans are very low-interest, and grants with no repayment requirement from federal
programs may share project costs. This results in interest rates in the 4% to 8% range. It is
common that a government agency will direct that all projects be evaluated at a specific rate.
As a matter of standardization, directives to use a specific interest rate are beneficial because
different government agencies are able to obtain varying types of funding at different rates.
This can result in projects of the same type being rejected in one state or city but accepted in
another. Standardized rates tend to increase the consistency of economic decisions and to reduce
gamesmanship.
The determination of the interest rate for public sector evaluation is as important as the determination
of the MARR for a private sector analysis. The public sector interest rate is identified
as i ; however, it is referred to by other names to distinguish it from the private sector rate. The
most common terms are discount rate and social discount rate.
Characteristic Public sector Private sector
Alternative selection Multiple criteria Primarily based on rate
criteria
of return
Multiple categories of users, economic as well as noneconomic interests, and special-interest
political and citizen groups make the selection of one alternative over another much more difficult
in public sector economics. Seldom is it possible to select an alternative on the sole basis of
a criterion such as PW or ROR. It is important to describe and itemize the criteria and selection
method prior to the analysis. This helps determine the perspective or viewpoint when the evaluation
is performed. Viewpoint is discussed below.
Characteristic Public sector Private sector
Environment of the evaluation Politically inclined Primarily economic
There are often public meetings and debates associated with public sector projects to accommodate
the various interests of citizens (owners). Elected officials commonly assist with the selection,
especially when pressure is brought to bear by voters, developers, environmentalists, and
others. The selection process is not as “clean” as in private sector evaluation.
The viewpoint of the public sector analysis must be determined before cost, benefit, and disbenefit
estimates are made and before the evaluation is formulated and performed. There are
several viewpoints for any situation, and the different perspectives may alter how a cash flow
estimate is classified.

9.1 Public Sector Projects 233
Some example perspectives are the citizen; the city tax base; number of students in the school district;
creation and retention of jobs; economic development potential; a particular industry interest
(agriculture, banking, electronics manufacturing); even the reelection of a public officeholder (often
termed pork projects ). In general, the viewpoint of the analysis should be as broadly defined as those
who will bear the costs of the project and reap its benefits. Once established, the viewpoint assists in
categorizing the costs, benefits, and disbenefits of each alternative, as illustrated in Example 9.1 .
EXAMPLE 9.1 Water Treatment Facility #3 Case
The situation with the location and construction of the new WTF3 and associated transmission
mains described in the chapter’s introduction has reached a serious level because of recent
questions posed by some city council members and citizen groups. Before going public to the
city council with the analysis performed last year, the director of Allen Water Utilities has
asked an engineering management consultant to review it and determine if it was an acceptable
analysis and correct economic decision, then and now. The lead consultant, Joel Whiterson,
took engineering economy as a part of his B.S. education and has previously worked on
economic studies in the government sector, but never as the lead person.
Within the first hour of checking background notes, Joel found several initial estimates
(shown below) from last year for expected consequences if WTF3 were built. He realized that
no viewpoint of the study was defined, and, in fact, the estimates were never classified as costs,
benefits, or disbenefits. He did determine that disbenefits were considered at some point in the
analysis, though the estimates for them are very sketchy.
Joel defined two viewpoints: a citizen of Allen and the Allen Water Utilities budget. He
wants to identify each of the estimates as a cost, benefit, or disbenefit from each viewpoint.
Please help with this classification.
Economic Dimension
1. Cost of water: 10% annual increase to Allen
households
2. Bonds: Annual debt service at 3% per year on
$540 million
3. Use of land: Payment to Parks and Recreation
for shaft sites and construction areas
4. Property values: Loss in value, sales price,
and property taxes
5. Water sales: Increases in sales to surrounding
communities
6. M&O: Annual maintenance and operations
costs
7. Peak load purchases: Savings in purchases of
treated water from secondary sources
Monetary Estimate
Average of $29.7 million (years 1–5, steady
thereafter)
$16.2 million (years 1–19); $516.2 million (year 20)
$300,000 (years 1–4)
$4 million (years 1–5)
$5 million (year 4) plus 5% per year (years 5–20)
$300,000 plus 4% per year increase (years 1–20)
$500,000 (years 5–20)
PE
Solution
The perspective of each viewpoint is identified and estimates are classified. (How this classification
is done will vary depending upon who does the analysis. This solution offers only one
logical answer.)
Viewpoint 1: Citizen of the city of Allen. Goal: Maximize the quality of life and wellness of
citizens with family and neighborhood as prime concerns.
Costs: 1, 2, 4, 6 Benefits: 5, 7 Disbenefits: 3
Viewpoint 2: Allen Water Utilities budget. Goal: Ensure the budget is balanced and of sufficient
size to fund rapidly growing city service demands.
Costs: 2, 3, 6 Benefits: 1, 5, 7 Disbenefits: 4
Citizens view costs in a different light than a city budget employee does. For example, the loss
of property values (item 4) is considered a real cost to a citizen, but is an unfortunate disbenefit

234 Chapter 9 Benefit/Cost Analysis and Public Sector Economics
from the city budget perspective. Similarly, the Allen Water Utilities budget interprets estimate
3 (payment for use of land to Parks and Recreation) as a real cost; but a citizen might interpret
this as merely a movement of funds between two municipal budgets—therefore, it is a disbenefit,
not a real cost.
Comment
The inclusion of disbenefits can easily change the economic decision. However, agreement on
the disbenefits and their monetary estimates is difficult (to impossible) to develop, often resulting
in the exclusion of any disbenefits from the economic analysis. Unfortunately, this usually
transfers the consideration of disbenefits to the noneconomic (i.e., political) realm of public
project decision making.
Most of the large public sector projects are developed through public-private partnerships
(PPPs). A partnership is advantageous in part because of the greater efficiency of the private sector
and in part because of the sizable cost to design, construct, and operate such projects. Full
funding by the government unit may not be possible using traditional means—fees, taxes, and
bonds. Some examples of the projects are as follows:
Project
Mass transportation
Bridges and tunnels
Ports and harbors
Airports
Water resources
Some Purposes of the Project
Reduce transit time; reduce congestion; improve environment; decrease road
accidents
Speed traffic flows; reduce congestion; improve safety
Increase cargo capacity; support industrial development; increase tourism
Increase capacity; improve passenger safety; support development
Desalination for drinking water; meet irrigation and industrial needs; improve
wastewater treatment
In these joint ventures, the public sector (government) is responsible for the funding and service
to the citizenry, and the private sector partner (corporation) is responsible for varying aspects of
the projects as detailed below. The government unit cannot make a profit, but the corporation(s)
involved can realize a reasonable profit; in fact, the profit margin is usually written into the contract
that governs the design, construction, and operation of the project.
Traditional methods of contracting were fi xed-price (historically called lump-sum) and cost
reimbursable (also called cost-plus). In these formats, a government unit took responsibility for
funding and possibly some of the design elements, and later all operation activities, while the
contractor did not share in the risks involved—liability, natural disasters, funding shortfalls, etc.
More recently, the PPP has become the arrangement of choice for most large public projects.
Commonly these are called design-build contracts, under which contractors take on more and
more of the functions from design to operation. Details are explained in publications such as
Design-Build Contracting Handbook (Cushman and Loulakis). The most reliance is placed upon
a contractor or contractors with a DBOMF contract, as described below.
The Design-Build-Operate-Maintain-Finance (DBOMF) contract is considered a turnkey approach
to a project. It requires the contractor(s) to perform all the DBOMF activities with collaboration
and approval of the owner (the government unit). The activity of financing is the
management of cash fl ow to support project implementation by a contracting firm. Although a
contractor may assist in some instances, the funding (obtaining the capital funds) remains the
government’s responsibility through bonding, commercial loans, taxation, grants, and gifts.
When the financing activity is not managed by a contractor, the contract is a DBOM; it is also
common to develop a design-build contract. In virtually all cases, some forms of design-build
arrangements for public projects are made because they offer several advantages to the government
and citizens served:
• Cost and time savings in the design, build, and operate phases
• Earlier and more reliable (less variable) cost estimates

9.2 Benefit/Cost Analysis of a Single Project 235
• Reduced administrative responsibilities for the owner
• Better efficiency of resource allocation of private enterprise
• Environmental, liability, and safety issues addressed by the private sector, where there usually
is greater expertise
Many of the projects in international settings and in developing countries utilize the publicprivate
partnership. There are, of course, disadvantages to this arrangement. One risk is that the
amount of funding committed to the project may not cover the actual build cost because it is
considerably higher than estimated. Another risk is that a reasonable profit may not be realized
by the private corporation due to low usage of the facility during the operate phase. To prevent
such problems, the original contract may provide for special subsidies and loans guaranteed by
the government unit. The subsidy may cover costs plus (contractually agreed-to) profit if usage
is lower than a specified level. The level used may be the breakeven point with the agreed-to
profit margin considered.
9.2 Benefit/Cost Analysis of a Single Project
The benefit/cost ratio is relied upon as a fundamental analysis method for public sector projects.
The B/C analysis was developed to introduce greater objectivity into public sector economics,
and as one response to the U.S. Congress approving the Flood Control Act of 1936. There are
several variations of the B/C ratio; however, the fundamental approach is the same. All cost and
benefit estimates must be converted to a common equivalent monetary unit (PW, AW, or FW) at
the discount rate (interest rate). The B/C ratio is then calculated using one of these relations:
B/C ———————
PW of benefits
———————
AW of benefits
———————
FW of benefits
PW of costs AW of costs FW of costs
[9.1]
Present worth and annual worth equivalencies are preferred to future worth values. The sign
convention for B/C analysis is positive signs; costs are preceded by a sign. Salvage values
and additional revenues to the government, when they are estimated, are subtracted from costs in
the denominator. Disbenefits are considered in different ways depending upon the model used.
Most commonly, disbenefits are subtracted from benefits and placed in the numerator. The
different formats are discussed below.
The decision guideline is simple:
If B/C 1.0, accept the project as economically justified for the estimates and discount rate
applied.
If B/C 1.0, the project is not economically acceptable.
Project evaluation
If the B/C value is exactly or very near 1.0, noneconomic factors will help make the decision.
The conventional B/C ratio, probably the most widely used, is calculated as follows:
B/C ——————————
benefits disbenefits
costs
B D ———
C
[9.2]
In Equation [9.2] disbenefits are subtracted from benefits, not added to costs. The B/C value
could change considerably if disbenefits are regarded as costs. For example, if the numbers
10, 8, and 5 are used to represent the PW of benefits, disbenefits, and costs, respectively, the
correct procedure results in B/C (10 8)5 0.40. The incorrect placement of disbenefits
in the denominator results in B/C 10(8 5) 0.77, which is approximately twice the correct
B/C value of 0.40. Clearly, then, the method by which disbenefits are handled affects the
magnitude of the B/C ratio. However, regardless of whether disbenefits are (correctly) subtracted
from the numerator or (incorrectly) added to costs in the denominator, a B/C ratio of
less than 1.0 by the first method will always yield a B/C ratio less than 1.0 by the second
method, and vice versa.
The modified B/C ratio includes all the estimates associated with the project, once operational.
Maintenance and operation (M&O) costs are placed in the numerator and treated in a manner

236 Chapter 9 Benefit/Cost Analysis and Public Sector Economics
similar to disbenefits. The denominator includes only the initial investment. Once all amounts are
expressed in PW, AW, or FW terms, the modified B/C ratio is calculated as
Modified B/C ————————————————
benefits disbenefits M&O costs
initial investment
[9.3]
Salvage value is usually included in the denominator as a negative cost. The modified B/C ratio
will obviously yield a different value than the conventional B/C method. However, as with disbenefits,
the modifi ed procedure can change the magnitude of the ratio but not the decision to
accept or reject the project.
The benefit and cost difference measure of worth, which does not involve a ratio, is based on
the difference between the PW, AW, or FW of benefits and costs, that is, B − C . If ( B − C ) 0,
the project is acceptable. This method has the advantage of eliminating the discrepancies noted
above when disbenefits are regarded as costs, because B represents net benefi ts. Thus, for the
numbers 10, 8, and 5, the same result is obtained regardless of how disbenefits are treated.
Subtracting disbenefits from benefits: B C (10 8) 5 3
Adding disbenefits to costs:
B C 10 (8 5) 3
Before calculating the B/C ratio by any formula, check whether the alternative with the larger
AW or PW of costs also yields a larger AW or PW of benefits. It is possible for one alternative
with larger costs to generate lower benefits than other alternatives, thus making it unnecessary to
further consider the larger-cost alternative.
By the very nature of benefits and especially disbenefits, monetary estimates are difficult to
make and will vary over a wide range. The extensive use of sensitivity analysis on the more
questionable parameters helps determine how sensitive the economic decision is to estimate
variation. This approach assists in determining the economic and public acceptance risk associated
with a defined project. Also, the use of sensitivity analysis can alleviate some of the public’s
concerns commonly expressed that people (managers, engineers, consultants, contractors,
and elected officials) designing (and promoting) the public project are narrowly receptive to different
approaches to serving the public’s interest.
EXAMPLE 9.2
In the past, the Afram Foundation has awarded many grants to improve the living and medical
conditions of people in war-torn and poverty-stricken countries throughout the world. In a
proposal for the foundation’s board of directors to construct a new hospital and medical clinic
complex in a deprived central African country, the project manager has developed some estimates.
These are developed, so she states, in a manner that does not have a major negative
effect on prime agricultural land or living areas for citizens.
Award amount: $20 million (end of) first year, decreasing by $5 million per year for 3
additional years; local government will fund during the first year only
Annual costs: $2 million per year for 10 years, as proposed
Benefits: Reduction of $8 million per year in health-related expenses for citizens
Disbenefits: $0.1 to $0.6 million per year for removal of arable land and commercial
districts
Use the conventional and modified B/C methods to determine if this grant proposal is economically
justified over a 10-year study period. The foundation’s discount rate is 6% per year.
Solution
Initially, determine the AW for each parameter over 10 years. In $1 million units,
Award:
Annual costs:
Benefits:
Disbenefits:
20 − 5(A/G,6%,4) $12.864 per year
$2 per year
$8 per year
Use $0.6 for the first analysis

9.2 Benefit/Cost Analysis of a Single Project 237
The conventional B/C analysis applies Equation [9.2].
B/C —————— 8.0 0.6
12.864 2.0 0.50
The modifi ed B/C analysis uses Equation [9.3].
Modified B/C ———————
8.0 0.6 2.0
0.42
12.864
The proposal is not justified economically since both measures are less than 1.0. If the low
disbenefits estimate of $0.1 million per year is used, the measures increase slightly, but not
enough to justify the proposal.
It is possible to develop a direct formula connection between the B/C of a public sector and
B/C of a private sector project that is a revenue alternative ; that is, both revenues and costs are
estimated. Further, we can identify a direct correspondence between the modified B/C relation in
Equation [9.3] and the PW method we have used repeatedly. (The following development also
applies to AW or FW values.) Let’s neglect the initial investment in year 0 for a moment, and
concentrate on the cash flows of the project for year 1 through its expected life. For the private
sector, the PW for project cash flows is
PW of project PW of revenue PW of costs
Since private sector revenues are approximately the same as public sector benefits minus disbenefits
(B D), the modified B/C relation in Equation [9.3] may be written as
Modified B/C
PW of (B D) PW of C
————————————
PW of initial investment
This relation can be slightly rewritten to form the profitability index (PI), which can be used to
evaluate revenue projects in the public or private sector. For years t 1, 2, . . . , n ,
PI
PW of NCF t
———————————
PW of initial investment
[9.4]
Note that the denominator includes only first cost (initial investment) items, while the numerator
has only cash flows that result from the project for years 1 through its life. The PI measure of
worth provides a sense of getting the most for the investment dollar (euro, yen, etc.). That is, the
result is in PW units per PW of money invested at the beginning. This is a “bang for the buck”
measure. When used solely for a private sector project, the disbenefits are usually omitted,
whereas they should be estimated and included in the modified B/C version of this measure for a
public project.
The evaluation guideline for a single project using the PI is the same as for the conventional
B/C or modified B/C.
If PI 1.0, the project is economically acceptable at the discount rate.
If PI 1.0, the project is not economically acceptable at the discount rate.
Remember, the computations for PI and modified B/C are essentially the same, except the PI is
usually applied without disbenefits estimated. The PI has another name: the present worth index
(PWI). It is often used to rank and assist in the selection of independent projects when the capital
budget is limited. This application is discussed in Chapter 12, Section 12.5.
Project evaluation
EXAMPLE 9.3
The Georgia Transportation Directorate is considering a public-private partnership with Young
Construction as the prime contractor using a DBOMF contract for a new 22.51-mile toll road on
the outskirts of Atlanta’s suburban area. The design includes three 4-mile-long commercial/
retail corridors on both sides of the toll road. Highway construction is expected to require

238 Chapter 9 Benefit/Cost Analysis and Public Sector Economics
5 years at an average cost of $3.91 million per mile. The discount rate is 4% per year, and the
study period is 30 years. Evaluate the economics of the proposal using (a) the modified B/C
analysis from the State of Georgia perspective and (b) the profitability index from the Young
corporate viewpoint in which disbenefits are not included.
Initial investment: $88 million distributed over 5 years; $4 million now and in year 5 and
$20 million in each of years 1 through 4.
Annual M&O cost: $1 million per year, plus an additional $3 million each fifth year, including
year 30.
Annual revenue/benefits: Include tolls and retail/commercial growth; start at $2 million in
year 1, increasing by a constant $0.5 million annually through year 10, and then increasing by
a constant $1 million per year through year 20 and remaining constant thereafter.
Estimable disbenefits: Include loss of business income, taxes, and property value in surrounding
areas; start at $10 million in year 1, decrease by $0.5 million per year through year
21, and remain at zero thereafter.
Solution
The PW values in year 0 for all estimates must be developed initially usually by hand, calculator,
or spreadsheet computations. If the 30 years of estimates are entered into a spreadsheet and
NPV functions at 4% are applied, the results in $1 million units are obtained. All values are
positive because of the sign convention for B/C and PI measures.
PW of investment $71.89 PW of benefits $167.41
PW of costs $26.87 PW of disbenefits $80.12
(a) From the public project perspective, the State will apply Equation [9.3].
Modified B/C ——————————
167.41 80.12 26.87
0.84
71.89
The toll road proposal is not economically acceptable, since B/C 1.0.
(b) From the private corporation viewpoint, Young Construction will apply Equation [9.4].
PI ———————
167.41 26.87
1.95
71.89
The proposal is clearly justified without the disbenefits, since PI > 1.0. The private project
perspective predicts that every investment dollar will return an equivalent of $1.95 over
30 years at a 4% per year discount rate.
Comment
The obvious question that arises concerns the correct measure to use. When PI is used in the private
project setting, there is no problem, since disbenefits are virtually never considered in the
economic analysis. The public project setting will commonly use some form of the B/C ratio with
the disbenefit considered. When a public-private partnership is initiated, there should be some
agreement beforehand that establishes the economic measure acceptable for analysis and decision
making throughout the project. Then the numerical dilemma presented above should not occur.
9.3 Alternative Selection Using Incremental
B/C Analysis
The technique to compare two mutually exclusive alternatives using benefit/cost analysis is virtually
the same as that for incremental ROR in Chapter 8. The incremental (conventional) B/C ratio,
which is identified as B/C, is determined using PW, AW, or FW calculations. The higher-cost
alternative is justified if B/C is equal to or larger than 1.0. The selection rule is as follows:
ME alternative
selection
If B/C 1.0, choose the higher-cost alternative, because its extra cost is economically justified.
If B/C 1.0, choose the lower-cost alternative.

9.3 Alternative Selection Using Incremental B/C Analysis 239
To perform a correct incremental B/C analysis, it is required that each alternative be compared
only with another alternative for which the incremental cost is already justified. This same rule
was used for incremental ROR analysis.
There are two dimensions of an incremental B/C analysis that differ from the incremental
ROR method in Chapter 8. We already know the first, all costs have a positive sign in the B/C
ratio. The second, and significantly more important, concerns the ordering of alternatives prior to
incremental analysis.
Alternatives are ordered by increasing equivalent total costs, that is, PW or AW of all cost
estimates that will be utilized in the denominator of the B/C ratio. When not done correctly, the
incremental B/C analysis may reject a justified higher-cost alternative.
If two alternatives, A and B, have equal initial investments and lives, but B has a larger equivalent
annual cost, then B must be incrementally justified against A. (This is illustrated in Example 9.4
below.) If this convention is not correctly followed, it is possible to get a negative cost value in
the denominator, which can incorrectly make B/C 1 and reject a higher-cost alternative that is
actually justified.
Follow these steps to correctly perform a conventional B/C ratio analysis of two alternatives.
Equivalent values can be expressed in PW, AW, or FW terms.
1. Determine the equivalent total costs for both alternatives.
2. Order the alternatives by equivalent total cost: first smaller, then larger. Calculate the incremental
cost (C) for the larger-cost alternative. This is the denominator in B/C.
3. Calculate the equivalent total benefits and any disbenefits estimated for both alternatives.
Calculate the incremental benefits (B) for the larger-cost alternative. This is (B D) if
disbenefits are considered.
4. Calculate the B/C ratio using Equation [9.2], (B D)/C.
5. Use the selection guideline to select the higher-cost alternative if B/C 1.0.
When the B/C ratio is determined for the lower-cost alternative, it is a comparison with the donothing
(DN) alternative. If B/C 1.0, then DN should be selected and compared to the second
alternative. If neither alternative has an acceptable B/C value and one of the alternatives does not
have to be selected, the DN alternative must be selected. In public sector analysis, the DN alternative
is usually the current condition.
EXAMPLE 9.4
The city of Garden Ridge, Florida, has received designs for a new patient room wing to the
municipal hospital from two architectural consultants. One of the two designs must be accepted
in order to announce it for construction bids. The costs and benefits are the same in most
categories, but the city financial manager decided that the estimates below should be considered
to determine which design to recommend at the city council meeting next week and to
present to the citizenry in preparation for an upcoming bond referendum next month.
Design A
Design B
Construction cost, $ 10,000,000 15,000,000
Building maintenance cost, $/year 35,000 55,000
Patient usage copay, $/year 450,000 200,000
The patient usage copay is an estimate of the amount paid by patients over the insurance coverage
generally allowed for a hospital room. The discount rate is 5%, and the life of the building is
estimated at 30 years.
(a) Use incremental B/C analysis to select design A or B.
(b) Once the two designs were publicized, the privately owned hospital in the directly adjacent
city of Forest Glen lodged a complaint that design A will reduce its own municipal hospital’s
income by an estimated $500,000 per year because some of the day-surgery features
of design A duplicate its services. Subsequently, the Garden Ridge merchants’ association
argued that design B could reduce its annual revenue by an estimated $400,000, because it

240 Chapter 9 Benefit/Cost Analysis and Public Sector Economics
will eliminate an entire parking lot used by their patrons for short-term parking. The city
financial manager stated that these concerns would be entered into the evaluation as disbenefits
of the respective designs. Redo the B/C analysis to determine if the economic decision
is still the same as when disbenefits were not considered.
Solution
( a) Since most of the cash flows are already annualized, the incremental B/C ratio will use
AW values. No disbenefit estimates are considered. Follow the steps of the procedure
above:
1. The AW of costs is the sum of construction and maintenance costs.
AW A 10,000,000(A/P,5%,30) 35,000 $685,500
AW B 15,000,000(A/P,5%,30) 55,000 $1,030,750
2. Design B has the larger AW of costs, so it is the alternative to be incrementally justified.
The incremental cost is
C AW B AW A $345,250 per year
3. The AW of benefits is derived from the patient usage copays, since these are consequences
to the public. The benefits for the B/C analysis are not the estimates themselves,
but the difference if design B is selected. The lower usage copay is a positive
benefit for design B.
B usage A usage B $450,000 $200,000 $250,000 per year
4. The incremental B/C ratio is calculated by Equation [9.2].
BC ————
$250,000
$345,250 0.72
5. The B/C ratio is less than 1.0, indicating that the extra costs associated with design B
are not justified. Therefore, design A is selected for the construction bid.
(b) The revenue loss estimates are considered disbenefits. Since the disbenefits of design B are
$100,000 less than those of A, this positive difference is added to the $250,000 benefits of
B to give it a total benefit of $350,000. Now
BC ————
$350,000
$345,250 1.01
Design B is slightly favored. In this case the inclusion of disbenefits has reversed the previous
economic decision. This has probably made the situation more difficult politically.
New disbenefits will surely be claimed in the near future by other special-interest groups.
Like other methods, incremental B/C analysis requires equal-service comparison of alternatives.
Usually, the expected useful life of a public project is long (25 or 30 or more years), so
alternatives generally have equal lives. However, when alternatives do have unequal lives, the
use of PW or AW to determine the equivalent costs and benefits requires that the LCM of lives be
used to calculate B/C. As with ROR analysis of two alternatives, this is an excellent opportunity
to use the AW equivalency of estimated (not incremental) costs and benefits, if the implied assumption
that the project could be repeated is reasonable. Therefore, use AW-based analysis of
actual costs and benefits for B/C ratios when different-life alternatives are compared.
EXAMPLE 9.5 Water Treatment Facility #3 Case
PE
As our case unfolds, the consultant, Joel Whiterson, has pieced together some of the B/C
analysis estimates for the 84-inch Jolleyville transmission main study completed last year. The
two options for constructing this main were open trench (OT) for the entire 6.8-mile distance
or a combination of trenching and bore tunneling (TT) for a shorter route of 6.3 miles. One of
the two options had to be selected to transport approximately 300 million gallons per day (gpd)
of treated water from the new WTF3 to an existing aboveground reservoir.

9.3 Alternative Selection Using Incremental B/C Analysis 241
The general manager of Allen Water Utilities has stated publicly several times that the
trench-tunnel combination option was selected over the open-trench alternative based on analysis
of both quantitative and nonquantitative data. He stated the equivalent annual costs in an
internal e-mail some months ago, based on the expected construction periods of 24 and
36 months, respectively, as equivalent to
AW OT $1.20 million per year
AW TT $2.37 million per year
This analysis indicated that the open-trench option was economically better, at that time. The
planning horizon for the transmission mains is 50 years; this is a reasonable study period, Joel
concluded. Use the estimates below that Joel has unearthed to perform a correct incremental
B/C analysis and comment on the results. The interest (discount) rate is 3% per year, compounded
annually, and 1 mile is 5280 feet.
Open trench (OT)
Trench-tunnel (TT)
Distance, miles 6.8 6.3
First cost, $ per foot 700 Trench for 2.0 miles: 700
Tunnel for 4.3 miles: 2100
Time to complete, months 24 36
Construction support costs, $ per month 250,000 175,000
Ancillary expenses, $ per month:
Environmental 150,000 20,000
Safety 140,000 60,000
Community interface 20,000 5,000
Solution
One of the alternatives must be selected, and the construction lives are unequal. Since it is not
reasonable to assume that this construction project will be repeated many cycles in the future,
it is incorrect to conduct an AW analysis over the respective completion periods of 24 and
36 months, or the LCM of these time periods. However, the study period of 50 years is a reasonable
evaluation time frame, since the mains are considered permanent installations. We can
assume that the construction first costs are a present worth value in year 0, but the equivalent
PW and 50-year AW of other monthly costs must be determined.
PW OT PW of construction PW of construction support costs
700(6.8)(5280) 250,000(12)( PA ,3%,2)
$30,873,300
AW OT 30,873,300( AP ,3%,50)
$1.20 million per year
PW TT [700(2.0) 2100(4.3)](5280) 175,000(12)( PA ,3%,3)
$61,010,460
AW TT 61,010,460( AP ,3%,50)
$2.37 million per year
The trench-tunnel (TT) alternative has a larger equivalent cost ; it must be justified against the
OT alternative. The incremental cost is
C AW TT AW OT 2.37 − 1.20 $1.17 million per year
The difference between ancillary expenses defines the incremental benefit for TT.
PW OT-anc
310,000(12)( PA ,3%,2)
$7,118,220
AW OT-anc 7,118,220( AP ,3%,50)
$276,685 per year

242 Chapter 9 Benefit/Cost Analysis and Public Sector Economics
PW TT-anc
85,000(12)( PA ,3%,3)
$2,885,172
AW TT-anc 2,885,172( AP ,3%,50)
$112,147 per year
B AW OT-anc AW TT-anc 276,685 112,147 $164,538 per year ($0.16 million)
Calculate the incremental B/C ratio.
B/C 0.16/1.17 0.14
Since B/C 1.0, the trench-tunnel option is not economically justified. Joel can now conclude
that the general manager’s earlier comment that the TT option was selected based on
quantitative and nonquantitative data must have had heavy dependence on nonquantitative
information not yet discovered.
9.4 Incremental B/C Analysis of Multiple, Mutually
Exclusive Alternatives
The procedure necessary to select one from three or more mutually exclusive alternatives using
incremental B/C analysis is essentially the same as that of Section 9.3. The procedure also parallels
that for incremental ROR analysis in Section 8.6. The selection guideline is as follows:
Choose the largest-cost alternative that is justified with an incremental B/C 1.0 when this selected
alternative has been compared with another justified alternative.
There are two types of benefit estimates—estimation of direct benefits, and implied benefits
based on usage cost estimates. The previous two examples (9.4 and 9.5) are good illustrations of
the second type of implied benefit estimation. When direct benefi ts are estimated, the B/C ratio
for each alternative may be calculated first as an initial screening mechanism to eliminate unacceptable
alternatives. At least one alternative must have B/C 1.0 to perform the incremental
B/C analysis. If all alternatives are unacceptable, the DN alternative is the choice. (This is the
same approach as that of step 2 for “revenue alternatives only” in the ROR procedure of Section
8.6. However, the term revenue alternative is not applicable to public sector projects.)
As in the previous section when comparing two alternatives, selection from multiple alternatives
by incremental B/C ratio utilizes equivalent total costs to initially order alternatives from
smallest to largest. Pairwise comparison is then undertaken. Also, remember that all costs are
considered positive in B/C calculations. The terms defender and challenger alternative are used
in this procedure, as in a ROR-based analysis. The procedure for incremental B/C analysis of
multiple alternatives is as follows:
1. Determine the equivalent total cost for all alternatives. Use AW, PW, or FW equivalencies.
2. Order the alternatives by equivalent total cost, smallest first.
3. Determine the equivalent total benefits (and any disbenefits estimated) for each alternative.
4. Direct benefi ts estimation only: Calculate the B/C for the first ordered alternative. If B/C
1.0, eliminate it. By comparing each alternative to DN in order, we eliminate all that have
B/C 1.0. The lowest-cost alternative with B/C 1.0 becomes the defender and the next
higher-cost alternative is the challenger in the next step. (For analysis by spreadsheet, determine
the B/C for all alternatives initially and retain only acceptable ones.)
5. Calculate incremental costs ( C ) and benefits ( B ) using the relations
C challenger cost defender cost
B challenger benefits defender benefits
If relative usage costs are estimated for each alternative, rather than direct benefits, B may
be found using the relation
B defender usage costs challenger usage costs [9.5]

9.4 Incremental B/C Analysis of Multiple, Mutually Exclusive Alternatives 243
6. Calculate the B/C for the first challenger compared to the defender.
B/C B C [9.6]
If B/C 1.0 in Equation [9.6], the challenger becomes the defender and the previous defender
is eliminated. Conversely, if B/C 1.0, remove the challenger and the defender
remains against the next challenger.
7. Repeat steps 5 and 6 until only one alternative remains. It is the selected one.
In all the steps above, incremental disbenefits may be considered by replacing B with ( B D ).
EXAMPLE 9.6
Schlitterbahn Waterparks of Texas, a very popular water and entertainment park headquartered
in New Braunfels, has been asked by four different cities outside of Texas to consider
building a park in their area. All the offers include some version of the following
incentives:
• Immediate cash incentive (year 0)
• A 10% of first-year incentive as a direct property tax reduction for 8 years
• Sales tax rebate sharing plan for 8 years
• Reduced entrance (usage) fees for area residents for 8 years
Table 9–1 (top section) summarizes the estimates for each proposal, including the present
worth of the initial construction cost and anticipated annual revenue. The annual M&O costs
are expected to be the same for all locations. Use incremental B/C analysis at 7% per year and
an 8-year study period to advise the board of directors if they should consider any of the offers
to be economically attractive.
Solution
The viewpoint is that of Schlitterbahn, and the benefits are direct estimates. Develop the AW
equivalents over 8 years, and use the procedure detailed above. The results are presented in
Table 9–1.
1. AW of total costs and an example for city 1 are determined in $1 million units.
AW of costs first cost(AP,7%,8) entrance fee reduction to residents
38.5(0.16747) 0.5
$6.948 ($6,948,000 per year)
2. The four alternatives are correctly ordered by increasing equivalent total cost in
Table 9–1.
TABLE 9–1 Incremental B/C Analysis of Water Park Proposals, Example 9.6
City 1 City 2 City 3 City 4
First cost, $ million 38.5 40.1 45.9 60.3
Entrance fee costs, $/year 500,000 450,000 425,000 250,000
Annual revenue, $ million/year 7.0 6.2 10.0 10.4
Initial cash incentive, $ 250,000 350,000 500,000 800,000
Property tax reduction, $/year 25,000 35,000 50,000 80,000
Sales tax sharing, $/year 310,000 320,000 320,000 340,000
AW of total costs, $ million/year 6.948 7.166 8.112 10.348
AW of total benefits, $ million/year 7.377 6.614 10.454 10.954
Overall B/C 1.06 0.92 1.29 1.06
Alternatives compared 1-to-DN B/C 1.0 3-to-1 4-to-3
Incremental costs C, $/year 6.948 1.164 2.236
Incremental benefits B, $/year 7.377 3.077 0.50
B/C 1.06 2.64 0.22
Increment justified? Yes Eliminated Yes No
City selected 1 3 3

244 Chapter 9 Benefit/Cost Analysis and Public Sector Economics
3. AW of total benefits and an example for city 1 are also determined in $1 million units.
AW of benefits revenue initial incentive(AP,7%,8)
property tax reduction sales tax sharing
7.0 0.25(0.16747) 0.025 0.31
$7.377 ($7,377,000 per year)
4. Since benefits are directly estimated (and no disbenefits are included), determine the overall
B/C for each alternative using Equation [9.1]. In the case of city 1,
BC 1 7.3776.948 1.06
City 2 is eliminated with B/C 2 0.92; the rest are initially acceptable.
5. The C and B values are the actual estimates for the 1-to-DN comparison.
6. The overall B/C is the same as B/C 1.06, using Equation [9.6]. City 1 is economically
justified and becomes the defender.
7. Repeat steps 5 and 6. Since city 2 is eliminated, the 3-to-1 comparison results in
C 8.112 6.948 1.164
B 10.454 7.377 3.077
BC 3.0771.164 2.64
City 3 is well justified and becomes the defender against city 4. From Table 9–1, B/C 0.22
for the 4-to-3 comparison. City 4 falls out easily, and city 3 is the one to recommend to the
board. Note that the DN alternative could have been selected had no proposal met the B/C or
B/C requirements.
Independent project
selection
When two or more independent projects are evaluated using B/C analysis and there is no
budget limitation, no incremental comparison is necessary. The only comparison is between
each project separately with the do-nothing alternative. The project B/C values are calculated,
and those with B/C 1.0 are accepted.
This is the same procedure as that used to select from independent projects using the ROR method
(Chapter 8). When a budget limitation is imposed, the capital budgeting procedure discussed in
Chapter 12 must be applied.
When the lives of mutually exclusive alternatives are so long that they can be considered
infinite, the capitalized cost is used to calculate the equivalent PW or AW values for costs
and benefits. Equation [5.3], A P ( i ), is used to determine the equivalent AW values in the
incremental B/C analysis. Example 9.7 illustrates this using the progressive example and a
spreadsheet.
EXAMPLE 9.7 Water Treatment Facility #3 Case
PE
Land for Water Treatment Facility #3 was initially purchased in the year 2010 for $19.3 million;
however, when it was publicized, influential people around Allen spoke strongly against
the location. We will call this location 1. Some of the plant design had already been completed
when the general manager announced that this site was not the best choice anyway, and that it
would be sold and a different, better site (location 2) would be purchased for $28.5 million.
This was well over the budget amount of $22.0 million previously set for land acquisition. As
it turns out, there was a third site (location 3) available for $35.0 million that was never seriously
considered.
In his review and after much resistance from Allen Water Utilities staff, the consultant, Joel,
received a copy of the estimated costs and benefits for the three plant location options. The
revenues, savings, and sale of bulk water rights to other communities are estimated as increments
from a base amount for all three locations. Using the assumption of a very long life for
the WTF3 facility and the established discount rate of 3% per year, determine what Joel discovered
when he did the B/C analysis. Was the general manager correct in concluding that
location 2 was the best, all said and done?

9.4 Incremental B/C Analysis of Multiple, Mutually Exclusive Alternatives 245
Location 1 Location 2 Location 3
Land cost, $ million 19.3 28.5 35.0
Facility first cost, $ million 460.0 446.0 446.0
Benefits, $ per year:
Pumping cost savings 5 3 0
Sales to area communities 12 10 8
Added revenue from Allen 6 6 6
Total benefits, $ per year 23 19 14
Solution
A spreadsheet can be very useful when performing an incremental B/C analysis of three or
more alternatives. Figure 9–1a presents the analysis with the preliminary input of AW values
for costs using the relation A P(i) and annual benefits. Figure 9–1b details all the functions
used in the analysis. Logical IF statements indicate alternative elimination and selection decisions.
In $1 million units,
Location 1: AW of costs A of land cost A of facility first cost
(19.3 460.0)(0.03)
$14.379 per year
AW of benefits $23
Location 2: AW of costs $14.235 AW of benefits $19
Location 3: AW of costs $14.430 AW of benefits $14
Though the AW of cost values are close to one another, the increasing order is locations 2, 1,
and 3 to determine B/C values. The benefits are direct estimates; therefore, the overall B/C
ratios indicate that location 3 (row 5; B/C 3 0.97) is not economically justified at the outset.
It is eliminated, and one of the remaining locations must be selected. Location 2 is justified
against the DN alternative (B/C 2 1.33); the only remaining comparison is 1-to-2 as detailed
in column C of Figure 9–1. Location 1 is a clear winner with B/C 27.78.
In conclusion, Joel has learned that location 1 is indeed the best and that, from the economic
perspective, the general manager was incorrect in stating that location 2 was better. However,
given the original evaluation criteria listed in the introduction—economics, environment, community
impact, and constructability—location 2 is likely a good compromise selection.
(a)
Figure 9–1
Incremental B/C analysis for WTF3 case: (a) numerical results and (b) functions developed for the analysis.
(b)

246 Chapter 9 Benefit/Cost Analysis and Public Sector Economics
Comment
This is an actual situation with changed names and values. Location 1 was initially purchased
and planned for WTF3. However, the presence of political, community, and environmental
stress factors changed the decision to location 2, when all was said and done.
9.5 Service Sector Projects and Cost-Effectiveness
Analysis
Much of the GDP of the United States and some countries in Europe and Asia is generated by
what has become known as the service sector of the economy. A large percentage of service sector
projects are generated by and dependent upon the private sector (corporations, businesses,
and other for-profit institutions). However, many projects in the public sector are also service
sector projects.
A service sector project is a process or system that provides services to individuals, businesses,
or government units. The economic value is developed primarily by the intangibles of the process
or system, not the physical entities , such as buildings, machines, and equipment. Manufacturing
and construction activities are commonly not considered a service sector project, though
they may support the theme of the service provided.
Service projects have a tremendous range of variety and purpose; to name a few: health care
systems, health and life insurance, airline reservation systems, credit card services, police and
court systems, security programs, safety training programs, and all types of consulting projects.
The intangible and intellectual work done by engineers and other professionals is often a part of
a service sector project.
The economic evaluation of a service project is difficult to a great degree because the cost and
benefit estimates are not accurate and often not within an acceptable degree of error. In other
words, undue risk may be introduced into the decision because of poor monetary estimates. For
example, consider the decision to place red-light cameras at stop lights to ticket drivers who run
the red light. This is a public and a service project, but its ( economic ) benefi ts are quite diffi cult
to estimate . Depending upon the viewpoint, benefits could be in terms of accidents averted,
deaths prevented, police personnel released from patrolling the intersection, or, from a more
mercenary viewpoint, amount of fines collected. In all but the last case, benefits in monetary
terms will be poor estimates. These are examples where B/C analysis does not work well and a
different form of analysis is needed.
In service and public sector projects, as expected, it is the benefits that are the more difficult
to estimate. An evaluation method that combines monetary cost estimates with nonmonetary
benefit estimates is cost-effectiveness analysis (CEA) . The CEA approach utilizes a costeffectiveness
measure or the cost-effectiveness ratio (CER) as a basis of ranking projects and
selecting the best of independent projects or mutually exclusive alternatives. The CER ratio is
defined as
equivalent total costs
CER —————————————
total of effectiveness measure — C E
[9.7]
In the red-light camera example, the effectiveness measure (the benefit) may be one of the samples
mentioned earlier, accidents averted or deaths prevented. Different from the B/C ratio of
costs to benefits, CER places the PW or AW of total costs in the numerator and the effectiveness
measure in the denominator. (The reciprocal of Equation [9.7] can also be used as the measure of
worth, but we will use CER as defined above.) With costs in the numerator, smaller ratio values
are more desirable for the same value of the denominator, since smaller ratio values indicate a
lower cost for the same level of effectiveness.
Like ROR and B/C analysis, cost-effectiveness analysis requires the ordering (ranking) of
alternatives prior to selection and the use of incremental analysis for mutually exclusive

9.5 Service Sector Projects and Cost-Effectiveness Analysis 247
alternative selection. Cost-effectiveness analysis utilizes a different ranking criterion than ROR or
BC analysis. The ordering criteria are as follows:
Independent projects: Initially rank projects by CER value.
Mutually exclusive alternatives: Initially rank alternatives by effectiveness measure, then
perform an incremental CER analysis.
Return again to the public/service project of red-light cameras. If the CER is defined as “cost per
total accidents averted ” and the projects are independent, increasing CER value is the ranking
basis. If the projects are mutually exclusive, “total accidents averted ” is the correct ranking basis
and an incremental analysis is necessary.
There are significantly different analysis procedures for independent and mutually exclusive
proposals. To select some from several (independent) projects, a budget limit, termed b, is inherently
necessary once ordering is complete. However, for selecting one from several (mutually
exclusive) alternatives, a pairwise incremental analysis is necessary and selection is made on the
basis of C/E ratios. The procedures and examples follow.
For independent projects, the procedure is as follows:
1. Determine the equivalent total cost C and effectiveness measure E , and calculate the CER
measure for each project.
2. Order projects from the smallest to the largest CER value.
3. Determine cumulative cost for each project and compare with the budget limit b .
4. The selection criterion is to fund all projects such that b is not exceeded.
Independent project
selection
EXAMPLE 9.8
Recent research indicates that corporations throughout the world need employees who demonstrate
creativity and innovation for new processes and products. One measure of these talents
is the number of patents approved each year through the R&D efforts of a company. Rollings
Foundation for Innovative Thinking has allocated $1 million in grant funds to award to corporations
that enroll their top R&D personnel in a 1- to 2-month professional training program in
their home state that has a historically proven track record over the last 5 years in helping
individuals earn patents.
Table 9–2 summarizes data for six corporations that submitted proposals. Columns 2 and 3
give the proposed number of attendees and cost per person, respectively, and column 4 provides
the historical track record of program graduates in patents per year. Use cost- effectiveness
analysis to select the corporations and programs to fund.
Solution
We assume that across all programs and all patent awards there is equal quality. Use the procedure
for independent projects and b $1 million to select from the proposals.
TABLE 9–2
Program
(1)
Data for Programs to Increase Patents Used for CEA
Total Personnel
(2)
Cost/Person, $
(3)
5-Year History,
Patents/Graduate/Year
(4)
1 50 5000 0.5
2 35 4500 3.1
3 57 8000 1.9
4 24 2500 2.1
5 12 5500 2.9
6 87 3800 0.6

248 Chapter 9 Benefit/Cost Analysis and Public Sector Economics
1. Using Equation [9.7], the effectiveness measure E is patents per year, and the CER is
program cost per person
CER ——————————
patents per graduate
C —
E
The program cost C is a PW value, and the E values are obtained from the proposals.
2. The CER values are shown in Table 9–3 in increasing order, column 5.
3. Cost per course, column 6, and cumulative costs, column 7, are determined.
4. Programs 4, 2, 5, 3, and 6 (68 of the 87 people) are selected to not exceed $1 million.
TABLE 9–3 Programs Ordered by CER Value, Example 9.8
Program
(1)
Total
Personnel
(2)
Cost/Person
C, $
(3)
Patents
per Year E
(4)
CER, $ per
Patent
(5) (3)/(4)
Program
Cost, $
(6) (2)(3)
Cumulative
Cost, $
(7) (6)
4 24 2,500 2.1 1,190 60,000 60,000
2 35 4,500 3.1 1,452 157,500 217,500
5 12 5,500 2.9 1,897 66,000 283,500
3 57 8,000 1.9 4,211 456,000 739,500
6 87 3,800 0.6 6,333 330,600 1,070,100
1 50 5,000 0.5 10,000 250,000 1,320,100
Comment
This is the first time that a budget limit has been imposed for the selection among independent
projects. This is often referred to as capital budgeting, which is discussed further in Chapter 12.
For mutually exclusive alternatives and no budget limit, the alternative with the highest effectiveness
measure E is selected without further analysis. Otherwise, an incremental CER analysis
is necessary and the budget limit is applied to the selected alternative(s). The analysis is
based on the incremental ratio C/E, and the procedure is similar to that we have applied for
incremental ROR and B/C, except now the concept of dominance is utilized.
Dominance occurs when the incremental analysis indicates that the challenger alternative offers
an improved incremental CER measure compared to the defender’s CER, that is,
ME alternative
selection
(C/E) challenger (C/E) defender
Otherwise, no dominance is present, and both alternatives remain in the analysis.
For mutually exclusive alternatives, the selection procedure is as follows:
1. Order the alternatives from smallest to largest effectiveness measure E . Record the cost for
each alternative.
2. Calculate the CER measure for the first alternative. This, in effect, makes DN the defender
and the first alternative the challenger. This CER is a baseline for the next incremental comparison,
and the first alternative becomes the new defender.
3. Calculate incremental costs ( C ) and effectiveness ( E ) and the incremental measure C/E
for the new challenger using the relation
cost of challenger cost of defender
C/E ———————————————————————
effectiveness of challenger effectiveness of defender —— C
E
4. If C/E C/E defender , the challenger dominates the defender and it becomes the new defender;
the previous defender is eliminated. Otherwise, no dominance is present and both
alternatives are retained for the next incremental evaluation.
5. Dominance present: Repeat steps 3 and 4 to compare the next ordered alternative (challenger)
and new defender. Determine if dominance is present.

9.5 Service Sector Projects and Cost-Effectiveness Analysis 249
Dominance not present: The current challenger becomes the new defender, and the next alternative
is the new challenger. Repeat steps 3 and 4 to compare the new challenger and new
defender. Determine if dominance is present.
6. Continue steps 3 through 5 until only one alternative or only nondominated alternatives remain.
7. Apply the budget limit (or other criteria) to determine which of the remaining alternative(s)
can be funded.
EXAMPLE 9.9
One of the corporations not selected for funding in Example 9.8 decided to fund its 50 R&D
personnel to attend one of the innovation and creativity programs at its own expense. One
criterion is that the program must have a historical average for a graduate of at least 2.0 patents
per year. Use the data in Table 9–2 to select the best program.
TABLE 9–4
Program
(1)
Total
Personnel
(2)
Mutually Exclusive Alternatives Evaluated by Cost-Effectiveness
Analysis, Example 9.9
Cost/Person
C , $
(3)
Patents per
Year E
(4)
CER, $ per
Patent
(5) (3)/(4)
Program Cost,
$
(6) (2)(3)
4 50 2,500 2.1 1190 125,000
5 50 5,500 2.9 1897 275,000
2 50 4,500 3.1 1452 225,000
Solution
From Table 9–2 , three programs—2, 4, and 5—have a historical record of at least two patents
per graduate per year. Since only one program will be selected, these are now mutually exclusive
alternatives. Use the procedure to perform the incremental analysis.
1. The alternatives are ranked by increasing patents per year in Table 9–4 , column 4.
2. The CER measure for program 4 is compared to the DN alternative.
program cost per person
C/E 4 —————————— ——— 2500
patents per graduate 2.1 1190
3. Program 5 is now the challenger.
5-to-4 comparison: C/E C ——
E
——————
5500 2500 3750
2.9 2.1
4. In comparison to C/E 4 1190, it costs $3750 per additional patent if program 5 is chosen
over 4. Program 5 is more expensive for more patents; however, clear dominance is not
present; both programs are retained for further evaluation.
5. Dominance not present: Program 5 becomes the new defender, and program 2 is the new
challenger. Perform the 2-to-5 comparison.
2-to-5 comparison: C/E C ——
E
——————
4500 5500 5000
3.1 2.9
Compared to C/E 5 1897, this increment is much cheaper—more patents for less money
per person. Dominance is present; eliminate program 5 and compare 2 to 4.
6. Repeat steps 3 through 5 and compare C/E to C/E 4 1190.
2-to-4 comparison: C/E C ——
E
——————
4500 2500 2000
3.1 2.1
This does not represent dominance of program 2 over 4. The conclusion is that both programs
are eligible for funding, that is, CEA in this case does not indicate only one program.
This occurs when there is not lower cost and higher effectiveness of one alternative
over another; that is, one alternative does not dominate all the others.
7. Now the budget and other considerations (probably noneconomic) are brought to bear to
make the final decision. The fact that program 4 costs $125,000—significantly less than
program 2 at $225,000—will likely enter into the decision.

250 Chapter 9 Benefit/Cost Analysis and Public Sector Economics
Cost-effectiveness analysis is a form of multiattribute decision-making in which economic
and noneconomic dimensions are integrated to evaluate alternatives from several perspectives by
different decision makers. See Chapter 10 for further discussion of alternative analysis using
multiple attributes.
9.6 Ethical Considerations in the Public Sector
Usual expectations by citizens of their elected officials—locally, nationally, and internationally—
are that they make decisions for the good of the public, ensuring safety and minimizing risk and cost
to the public. Above these is the long-standing expectation that public servants have integrity.
Similarly, the expectations of engineers employed by government departments, and those
serving as consultants to government agencies, are held to high standards. Impartiality, consideration
of a wide range of circumstances, and the use of realistic assumptions are but three of
the foundation elements upon which engineers should base their recommendations to decision
makers. This implies that engineers in public service avoid
• Self-serving, often greedy individuals and clients with goals of excessive profits and future
contract awards
• Using a politically favorable perspective that compromises the results of a study
• Narrowly-defined assumptions that serve special interest groups and subcommunities potentially
affected by the findings
Many people are disappointed and discouraged with government when elected officials and
public employees (engineers and others, alike) do not have real commitment to integrity and
unbiasedness in their work.
Engineers are routinely involved in two of the major aspects of public sector activities:
Public policy making —the development of strategy for public service, behavior, fairness,
and justice. This may involve literature study, background discovery, data collection, opinion
giving, and hypothesis testing. An example is transportation management . Engineers make
virtually all the recommendations based on data and long-standing decision algorithms for
policy items such as capacity of roads, expansion of highways, planning and zoning rules,
traffic signal usage, speed limit corridors, and many related topics in transportation policy.
Public officials use these findings to establish public transportation policy.
Public planning —the development of projects that implement strategy and affect people,
the environment, and financial resources in a variety of ways. Consider traffi c control ,
where the use and placement of traffic control signs, signals, speed limits, parking restrictions,
etc. are detailed based upon established policy and current data. (In effect, this is
systems engineering, that is, an application of the life-cycle phases and stages explained in
Section 6.5.)
Whether in the arena of policy making or public planning, engineers can find ethical compromise
a possibility when working with the public sector. A few circumstances are summarized here.
• Use of technology Many public projects involve the use of new technology. The public risks
and safety factors are not always known for these new advances. It is common and expected
that engineers make every attempt to apply the latest technology while ensuring that the public
is not exposed to undue risk.
• Scope of study A client may pressure the engineers to limit the range of options, the assumption
base, or the breadth of alternative solutions. These restrictions may be based on financial
reasons, politically-charged topics, client-favored options, or a wide variety of other reasons.
To remain impartial, it is the responsibility of the engineer to submit a fully unbiased analysis,
report, and recommendation, even though it may jeopardize future contract possibilities, promote
public disfavor, or generate other negative consequences.
• Negative community impact It is inevitable that public projects will adversely affect some
groups of people, or the environment, or businesses. The intentional silencing of these projected
effects is often the cause of strong public outcries against what may be a project that is
in the best interest of the community at large. Engineers who find (stumble onto) such negative
impacts may be pressured by clients, managers, or public figures to overlook them, though

Chapter Summary 251
TABLE 9–5 Some Ethical Considerations When Performing B/C and CEA Analysis
What the Study Includes Ethical Dimension
Audience for study
Impact time of decision
Greater good for community
as a whole
Reliance on economic
measures only
Scope of disbenefits estimated
and evaluated
Is it ethical to select a specific
group of people affected by the
project and neglect possible
effects on other groups?
Is it ethical to decide now for future
generations who may be adversely
and economically affected
by the current project decision?
Vulnerable minority groups, especially
economically deprived
ones, may be disproportionally
affected. Is this ethical if the impact
is predictable?
Is it acceptable to reduce all
costs and benefits to monetary
estimates for a decision, then
subjectively impute nonquantified
factors in the final decision?
Is it ethical to disregard any disbenefits
in the B/C study or use
indirect effectiveness measures
in a CEA study based on the difficulty
to estimate some of them?
Example
Construct children’s health care clinics
for city dwellers, but neglect rural families
with poor transportation means.
Accomplish financial bailouts of corporations
when future generations’ taxes
will be significantly higher to recover the
costs, plus interest and inflation effects.
Allow a chemical plant that is vital to
the community’s employment to pollute
a waterway when a minority group is
known to eat fish from the water that is
predictably contaminated.
Softening of building codes can improve
the financial outlook for home
builders; however, increased risks of
fire loss, storm and water damage to
structures, and reduced future resale
values are considered only in passing as
a new subdivision is approved by the
planning and zoning committee.
Noise and air pollution caused by a
planned open-pit quarry will have a negative
effect on area ranchers, residents,
wildlife, and plant life; but the effectiveness
measure considers only suburban
residents due to estimation difficulty of
effects on other constituencies.
the Code of Ethics for Engineers dictates a full and fair analysis and report. For example, a
planned rerouting of a city street may effectively cut off a section of international citizens’
businesses, thus resulting in a clearly predictable economic downturn. Considering this outcome
in the recommendation to the transportation department should be a goal of the analyzing
engineers, yet pressure to bias the results may be quite high.
The results of a B/C or CEA analysis are routinely depended upon by public officials and
government staff members to assist in making public planning decisions. As discussed earlier,
estimations for benefits, disbenefits, effectiveness measures, and costs can be difficult and inaccurate,
but these analysis tools are often the best available to structure a study. Some example
ethically-oriented challenges that may be confronted during B/C and CEA analyses are summarized
in Table 9–5 .
CHAPTER SUMMARY
The benefit/cost method is used primarily to evaluate alternatives in the public sector. When one
is comparing mutually exclusive alternatives, the incremental B/C ratio must be greater than or
equal to 1.0 for the incremental equivalent total cost to be economically justified. The PW, AW,
or FW of the initial costs and estimated benefits can be used to perform an incremental B/C
analysis. For independent projects, no incremental B/C analysis is necessary. All projects with
B/C 1.0 are selected provided there is no budget limitation. It is usually quite diffi cult to make
accurate estimates of benefi ts for public sector projects. The characteristics of public sector projects
are substantially different from those of the private sector: initial costs are larger; expected
life is longer; additional sources of capital funds include taxation, user fees, and government
grants; and interest (discount) rates are lower.

252 Chapter 9 Benefit/Cost Analysis and Public Sector Economics
Service projects develop economic value largely based on the intangibles of the services provided
to users, not the physical items associated with the process or system. Evaluation by B/C
analysis can be difficult with no good way to make monetary estimates of benefits. Cost-effectiveness
analysis (CEA) combines cost estimates and a nonmonetary effectiveness measure (the
benefit) to evaluate independent or mutually exclusive projects using procedures that are similar
to incremental ROR and B/C analysis. The concept of dominance is incorporated into the procedure
for comparing mutually exclusive alternatives.
As a complement to the discussion on professional ethics in Chapter 1, some potential ethical
challenges in the public sector for engineers, elected officials, and government consultants are
discussed here. Examples are included.
PROBLEMS
Public Projects and B/C Concepts
9.1 What is the difference between disbenefits and
costs?
9.2 Identify the following as primarily public or private
sector undertakings: eBay, farmer’s market,
state police department, car racing facility, social
security, EMS, ATM, travel agency, amusement
park, gambling casino, swap meet, football
stadium.
9.3 State whether the following characteristics are primarily
associated with public or private sector
projects: large initial investment, park user fees,
short-life projects, profit, disbenefits, tax-free
bonds, subsidized loans, low interest rate, income
tax, water quality regulations.
9.4 Identify the following cash flows as a benefit, disbenefit,
or cost.
(a) Loss of income to local businesses because
of a new freeway
(b) Less travel time because of a loop bypass
(c) $400,000 annual income to local businesses
because of tourism created by a national park
(d) Cost of fish from a hatchery to stock a lake at
the state park
(e) Less tire wear because of smoother road
(f)
surfaces
Decrease in property values due to the closure
of a government research lab
(g) School overcrowding because of a military
base expansion
(h) Revenue to local motels because of an extended
weekend holiday
9.5 What is a fundamental difference between DBOM
and DBOMF contracts?
9.6 Buster County has proposed a strict water conservation
policy for all industrial plants within the
county limits. Enforcement is proposed to be a
public-private partnership between the sheriff’s
office and a private security company. In preparing
for a B/C analysis of the proposal, estimates must
be categorized as a benefit, cost, or disbenefit. The
categories chosen will vary depending upon a person’s
viewpoint. Assume that possible viewpoints
include sales revenues, politics, service to the
public, the environment, contract obligations, future
generations of residents, legal matters, revenue
and budget (in general), and customers.
( a) Select the top two viewpoints (in your opinion)
for each of the following individuals as
they would categorize estimates as a cost,
benefit, or disbenefit.
1. An industrial plant manager in the county
2. County sheriff’s deputy (appointed office)
3. County commissioner (elected office)
4. Security company president
( b) Explain your answers by writing a short description
of why you selected as you did.
Project B/C Value
9.7 If an alternative has a salvage value, how is it handled
in the calculation of a B/C ratio relative to
benefits, disbenefits, costs, or savings?
9.8 The cost of grading and spreading gravel on a
short rural road is expected to be $300,000. The
road will have to be maintained at a cost of $25,000
per year. Even though the new road is not very
smooth, it allows access to an area that previously
could only be reached with off-road vehicles. The
improved accessibility has led to a 150% increase
in the property values along the road. If the previous
market value of a property was $900,000,
calculate the B/C ratio using an interest rate of 6%
per year and a 20-year study period.
9.9 Arsenic enters drinking water supplies from natural
deposits in the earth or from agricultural and
industrial practices. Since it has been linked to

Problems 253
cancer of the bladder, kidney, and other internal
organs, the EPA has lowered the arsenic standard
for drinking water from 0.050 parts per million to
0.010 parts per million (10 parts per billion). The
annual cost to public water utilities to meet the
new standard is estimated to be $200 per household.
If it is estimated that there are 90 million
households in the United States and that the lower
standard can save 50 lives per year valued at
$4,000,000 per life, what is the benefit/cost ratio of
the regulation?
9.10 From the following estimates, determine the B/C
ratio for a project that has a 20-year life. Use an
interest rate of 8% per year.
Consequences to
the People
Annual
benefits
$90,000
per year
Annual $10,000
disbenefits per year
Consequences to
the Government
First
cost
Annual
cost
Annual
savings
$750,000
$50,000
per year
$30,000
per year
9.11 A project to extend irrigation canals into an area
that was recently cleared of mesquite trees (a nuisance
tree in Texas) and large weeds is projected to
have a capital cost of $2,000,000. Annual maintenance
and operation costs will be $100,000 per
year. Annual favorable consequences to the general
public of $820,000 per year will be offset to
some extent by annual adverse consequences of
$400,000 to a portion of the general public. If the
project is assumed to have a 20-year life, what is
the B/C ratio at an interest rate of 8% per year?
9.12 Calculate the B/C ratio for the following cash flow
estimates at a discount rate of 7% per year.
Item
Cash Flow
FW of benefits, $ 30,800,000
AW of disbenefits, $ per year 105,000
First cost, $ 1,200,000
M&O costs, $ per year 400,000
Life of project, years 20
9.13 The benefits associated with a nuclear power plant
cooling water filtration project located on the Ohio
River are $10,000 per year forever, starting in
year 1. The costs are $50,000 in year 0 and $50,000
at the end of year 2. Calculate the B/C ratio at i
10% per year.
9.14 A privately funded wind-based electric power
generation company in the southern part of the
country has developed the following estimates
(in $1000) for a new turbine farm. The MARR is
10% per year, and the project life is 25 years. Calculate
( a ) the profitability index and ( b ) the modified
B/C ratio.
Benefits: $20,000 in year 0 and $30,000 in
year 5
Savings: $2000 in years 1–20
Cost: $50,000 in year 0
Disbenefits: $3000 in years 1–10
9.15 For the values shown, calculate the conventional
B/C ratio at i 10% per year.
PW, $ AW, $/Year FW, $
First cost 100,000 — 259,370
M&O cost 61,446 10,000 159,374
Benefits — 40,000 637,496
Disbenefits 30,723 5,000 —
9.16 A proposal to reduce traffic congestion on I-5 has
a B/C ratio of 1.4. The annual worth of benefits
minus disbenefits is $560,000. What is the first
cost of the project if the interest rate is 6% per year
and the project is expected to have a 20-year life?
9.17 Oil spills in the Gulf of Mexico have been known
to cause extensive damage to both public and private
oyster grounds along the Louisiana and Mississippi
shores. One way to protect shellfish along
the shoreline is to release large volumes of freshwater
from the Mississippi River to flush oil out to
sea. This procedure inevitably results in death to
some of the saltwater shellfish while preventing
more widespread destruction to public reefs. Oil
containment booms and other temporary structures
can also be used to intercept floating oil before
it damages sensitive fishing grounds. If the
Fish and Wildlife Service spent $110 million in
year 0 and $50 million in years 1 and 2 to minimize
environmental damage from one particular
oil spill, what is the benefit-to-cost ratio provided
the efforts resulted in saving 3000 jobs valued at a
total of $175 million per year? Assume disbenefits
associated with oyster deaths amounted to $30 million
in year 0. Use a 5-year study period and an
interest rate of 8% per year.
9.18 On-site granular ferric hydroxide (GFH) systems
can be used to remove arsenic from water when
daily flow rates are relatively low. If the operating
cost is $600,000 per year and the public health
benefits are assumed to be $800,000 per year, what
initial investment in the GFH system is necessary
to guarantee a modified B/C ratio of at least 1.0?
Assume that the equipment life is 10 years and the
interest rate is 6% per year.

254 Chapter 9 Benefit/Cost Analysis and Public Sector Economics
9.19 The Parks and Recreation Department of Burkett
County has estimated that the initial cost of a
“bare-bones” permanent river park will be
$2.3 million. Annual upkeep costs are estimated at
$120,000. Benefits of $340,000 per year and disbenefits
of $40,000 per year have also been identified.
Using a discount rate of 6% per year, calculate
( a ) the conventional B/C ratio and ( b ) the
modified B/C ratio.
9.20 From the following data, calculate the ( a ) conventional
and ( b ) modified benefit/cost ratios using an
interest rate of 6% per year and an infinite project
period.
Benefits:
To the
People
$300,000 now and
$100,000 per
year thereafter
Disbenefits: $40,000 per year
To the
Government
Costs: $1.5 million now
and $200,000
three years
from now
Savings: $70,000 per year
9.21 In 2010, Brazil began construction of the Belo
Monte hydroelectric dam on the Xingu River
(which feeds the Amazon River). The project is
funded by a consortium of investors and is expected
to cost $11 billion. It will begin producing
electricity in 2015. Even though the dam will provide
clean energy for millions of people, environmentalists
are sharply opposed. They say it will
devastate wildlife and the livelihoods of 40,000
people who live in the area to be flooded.
Assume that the funding will occur evenly over
the 5-year period from 2010 through 2014 at
$2.2 billion per year. The disbenefits are estimated
to be $100,000 for each displaced person and $1
billion for wildlife destruction. Assume that the
disbenefits will occur evenly through the 5-year
construction period and anticipated benefits will
begin at the end of 2015 and continue indefinitely.
Use an interest rate of 8% per year to determine
what the equivalent annual benefits must be to ensure
a B/C ratio of at least 1.0.
9.22 In the United States, the average number of airplanes
in the sky on an average morning is 4000.
There are another 16,000 planes on the ground.
Aerospace company Rockwell Collins developed
what it calls a digital parachute —a panic-button
technology that will land any plane in a pinch at
the closest airport, no matter what the weather or
geography and without the help of a pilot. The
technology can be applied if a pilot is no longer
capable of flying the plane or is panicked and confused
about what to do in an emergency. Assume
that the cost of retrofitting 20,000 commercial airplanes
is $100,000 each and the plane stays in service
for 15 years. If the technology saves an average
of 30 lives per year, with the value of a human
life placed at $4,000,000, what is the B/C ratio?
Use an interest rate of 10% per year.
9.23 Although the lower Rio Grande is regulated by
the Elephant Butte Dam and the Caballo Reservoir,
serious flooding has occurred in El Paso and
other cities located along the river. This has required
homeowners living in valley areas near the
river to purchase flood insurance costing between
$145 and $2766 per year. To alleviate the possibility
of flooding, the International Boundary and
Water Commission undertook a project costing
$220 million to raise the levees along flood-prone
portions of the river. As a result, 13,000 properties
were freed of the federal mandate to purchase
flood insurance. In addition, historical records indicate
that damage to infrastructure will be
avoided, which amounts to an average benefit of
$8,200,000 per year. If the average cost of flood
insurance is $460 per household per year, calculate
the benefit-to-cost ratio of the levee-raising
project. Use an interest rate of 6% per year and a
30-year study period.
9.24 For the data shown, calculate the conventional B/C
ratio at i 6% per year.
Benefits:
$20,000 in year 0 and
$30,000 in year 5
Disbenefits: $7000 in year 3
Savings (to government): $25,000 in years 1–4
Cost: $100,000 in year 0
Project life:
5 years
9.25 Explain a fundamental difference between the
modified B/C ratio and the profitability index.
9.26 Gerald Corporation entered a public-private partnership
using a DBOM contract with the state of
Massachusetts 10 years ago for railroad system
upgrades. Determine the profitability index for the
financial results listed below using a MARR of
8% per year.
Investments: Year 0 $4.2 million
Year 5 $3.5 million
Net savings: Years 15 $1.2 million per year
Years 610 $2.5 million per year
9.27 A project had a staged investment distributed over
the 6-year contract period. For the cash flows
shown (next page) and an interest rate of 10% per
year, determine the profitability index and determine
if the project was economically justified.

Problems 255
Year 0 1 2 3 4 5 6
Investment, $1000 25 0 10 0 5 0 0
NCF, $1000 per year 0 5 7 9 11 13 15
Two Alternative Comparison
9.28 In comparing two alternatives by the B/C method,
if the overall B/C ratio for both alternatives is calculated
to be exactly 1.0, which alternative should
you select?
9.29 In comparing alternatives X and Y by the B/C
method, if B/C X 1.6 and B/C Y 1.8, what is
known about the B/C ratio on the increment of investment
between X and Y?
9.30 Logan Well Services Group is considering two sites
for storage and recovery of reclaimed water. The
mountain site (MS) will use injection wells that cost
$4.2 million to develop and $280,000 per year for
M&O costs. This site will be able to accommodate
150 million gallons per year. The valley site (VS) will
involve recharge basins that cost $11 million to construct
and $400,000 to operate and maintain. At this
site, 890 million gallons can be injected each year. If
the value of the injected water is $3.00 per 1000 gallons,
which alternative, if either, should be selected
according to the B/C ratio method? Use an interest
rate of 8% per year and a 20-year study period.
9.31 The estimates shown are for a bridge under consideration
for a river crossing in Wheeling, West
Virginia. Use the B/C ratio method at an interest
rate of 6% per year to determine which bridge, if
either, should be built.
East Location
West Location
Initial cost, $ 11 10 6 27 10 6
Annual M&O, $/year 100,000 90,000
Benefits, $/year 990,000 2,400,000
Disbenefits, $/year 120,000 100,000
Life, years
9.32 Select the better of two proposals to improve street
safety and lighting in a colonia in south central
New Mexico. Use a B/C analysis and an interest
rate of 8% per year.
Proposal 1 Proposal 2
Initial cost, $ 900,000 1,700,000
Annual M&O cost, $/year 120,000 60,000
Annual benefits, $/year 530,000 650,000
Annual disbenefits, $/year 300,000 195,000
Life, years 10 20
9.33 Conventional and solar alternatives are available
for providing energy at a remote radar site. Use the
B/C ratio to determine which method should be
selected at an interest rate of 8% per year and a
5-year study period.
Conventional
Solar
Initial cost, $ 300,000 2,500,000
Annual cost, $ per year 700,000 5,000
9.34 The two alternatives shown are under consideration
for improving security at a county jail in Travis
County, New York. Determine which one should be
selected, based on a B/C analysis, an interest rate of
7% per year and a 10-year study period.
Extra Cameras
(EC)
New Sensors
(NS)
First cost, $ 38,000 87,000
Annual M&O, $/year 49,000 64,000
Benefits, $/year 110,000 160,000
Disbenefits, $/year 26,000 —
9.35 The U.S. government recently released an RFP to
construct a second-story floor on an existing building
at the Pentagon Complex. Separate contractors proposed
two methods. Method 1 will use lightweight
expanded shale on a metal deck with open web joists
and steel beams. For this method, the costs will be
$14,100 for concrete, $6000 for metal decking,
$4300 for joists, and $2600 for beams. Method 2 will
construct a reinforced concrete slab costing $5200
for concrete, $1400 for rebar, $2600 for equipment
rental, and $1200 for expendable supplies. Special
additives will be included in the lightweight concrete
that will improve the heat-transfer properties of the
floor. If the energy costs for method 1 will be $600
per year lower than for method 2, which one is more
attractive? Use an interest rate of 7% per year, a
20 -year study period, and the B/C method.
9.36 A project to control flooding from rare, but sometimes
heavy rainfalls in the arid southwest will
have the cash flows shown below. Determine
which project should be selected on the basis of a
B/C analysis at i 8% per year and a 20-year
study period.
Sanitary
Sewers
Open
Channels
First cost, $ 26 million 53 million
M&O cost, $ per year 400,000 30,000
Homeowner cleanup costs,
$ per year
60,000 0
9.37 Two routes are under consideration for a new interstate
highway. The long intervalley route would be
25 kilometers in length and would have an initial
cost of $25 million. The short transmountain route

256 Chapter 9 Benefit/Cost Analysis and Public Sector Economics
would be 10 kilometers long and would have an initial
cost of $45 million. Maintenance costs are estimated
at $150,000 per year for the long route and
$35,000 per year for the short route. Regardless of
which route is selected, the volume of traffic is expected
to be 400,000 vehicles per year. If the vehicle
operating expenses are assumed to be $0.30 per kilometer,
determine which route should be selected on
the basis of ( a ) conventional B/C analysis and
( b ) modified B/C analysis. Assume an infinite life for
each road, and use an interest rate of 8% per year.
9.38 The Idaho Department of Fish and Wildlife (IDFW)
is considering two locations for a new state park.
Location E would require an investment of $3 million
and $50,000 per year to maintain. Location W
would cost $7 million to construct, but the IDFW
would receive an additional $25,000 per year in
park fees. The operating cost of location W will be
$65,000 per year. The revenue to park concessionaires
will be $500,000 per year at location E and
$700,000 at location W. The disbenefits associated
with each location are $30,000 per year for location
E and $40,000 per year for location W. Assume the
park will be maintained indefinitely. Use an interest
rate of 12% per year to determine which location, if
either, should be selected on the basis of ( a ) the
B/C method and ( b ) the modified B/C method.
9.39 Three engineers made the estimates shown below
for two optional methods by which new construction
technology would be implemented at a site for public
housing. Either one of the two options or the current
method may be selected. Set up a spreadsheet
for B/C sensitivity analysis and determine if option
1, option 2 or the do-nothing option is selected by
each of the three engineers. Use a life of 5 years and
a discount rate of 10% per year for all analyses.
Engineer Bob Engineer Judy Engineer Chen
Option 1 Option 2 Option 1 Option 2 Option 1 Option 2
Alternative A B C D E F
PW of capital, $ 80 50 72 43 89 81
PW of benefits, $ 70 55 76 52 85 84
9.41 Comparison of five mutually exclusive alternatives
is shown. One must be accepted. According
to the B/C ratio, which alternative should be selected
(costs increase from A to E).
Comparison
B/C
Ratio
A versus B 0.75
B versus C 1.4
C versus D 1.3
A versus C 1.1
A versus D 0.2
B versus D 1.9
C versus E 1.2
D versus E 0.9
9.42 A consulting engineer is currently evaluating four
different projects for the Department of Housing
and Urban Development. The future worth of
costs, benefits, disbenefits, and cost savings is
shown. The interest rate is 10% per year, compounded
continuously. Determine which of the
projects, if any, should be selected, if the projects
are ( a ) independent and ( b ) mutually exclusive.
Project ID
Good Better Best Best of All
FW of first costs, $ 10,000 8,000 20,000 14,000
FW of benefits, $ 15,000 11,000 25,000 42,000
FW of disbenefits, $ 6,000 1,000 20,000 32,000
FW of cost savings, $ 1,500 2,000 16,000 3,000
9.43 From the data shown below for six mutually
exclusive projects, determine which project, if
any, should be selected.
Project ID
Initial cost, $ 50,000 90,000 75,000 90,000 60,000 70,000
Cost, $/year 3,000 4,000 3,800 3,000 6,000 3,000
Benefits, $/year 20,000 29,000 30,000 35,000 30,000 35,000
Disbenefits, $/year 500 1,500 1,000 0 5,000 1,000
Multiple Alternatives
9.40 A group of engineers responsible for developing
advanced missile detection and tracking technologies,
such as shortwave infrared, thermal infrared
detection, target tracking radar, etc., recently came
up with six proposals for consideration. The present
worth (in $ billions) of the capital requirements
and benefits is shown for each alternative in the
table. Determine which one(s) should be undertaken,
if they are ( a ) independent and ( b ) mutually
exclusive.
Annual cost, $
per year
Annual benefits,
$ per year
B/C ratio (alternative
vs. DN)
A B C D E F
8000 25,000 15,000 32,000 17,000 20,000
? ? ? ? ? ?
1.23 1.12 0.87 0.97 0.71 1.10
Selected Incremental B/C Ratios
A versus B 1.07
A versus C 0.46
A versus F 1.02
B versus D 0.43
B versus E 2.00
B versus F 1.20
C versus D 1.06
C versus F 1.80

Problems 257
9.44 Four mutually exclusive revenue alternatives are
being compared using the B/C method. Which alternative,
if any, should be selected?
Alternative
Initial
Cost, $ Millions
B/C Ratio
vs. DN
Incremental B/C
When Compared
with Alternative
A B C D
A 30 0.87 — 2.38 1.30 1.38
B 38 1.18 — 0.58 1.13
C 52 1.04 — 1.45
D 81 1.16 —
9.45 The city of St. Louis, Missouri, is considering various
proposals regarding the disposal of used tires.
All of the proposals involve shredding, but the
charges for the service and the handling of the tire
shreds differ in each plan. An incremental B/C
analysis was initiated but never completed.
( a) Fill in all the missing blanks in the table.
(b) Determine which alternative should be selected.
Alternative
PW of
Costs,$
PW of
Benefits, $
PW of
Disbenefits, $
B/C
Ratio
Incremental
B/C When
Compared with
Alternative
J K L M
J 20 ? 1 1.05 — ? ? ?
K 23 28 ? 1.13 — ? ?
L 28 35 3 ? — ?
M ? 51 4 1.34 —
Cost-Effectiveness Analysis
9.46 In an effort to improve productivity in a large
semiconductor manufacturing plant, the plant
manager decided to undertake on a trial basis a series
of actions directed toward improving employee
morale. Six different strategies were implemented,
such as increased employee autonomy,
flexible work schedules, improved training, company
picnics, electronic suggestion box, and better
work environment. Periodically, the company surveyed
the employees to measure the change in morale.
The measure of effectiveness is the difference
between the number of employees who rate their
job satisfaction as very high and those who rate it
very low. The per-employee cost of each strategy
(identified as A through F) and the resultant measurement
score are shown in the next column.
The manager has a maximum of $50 per employee
to spend on the permanent implementation of as
many of the strategies as are justified from both the
effectiveness and economic viewpoints. Determine
which strategies are the best to implement.
(Hand or spreadsheet solution is acceptable, as you
are instructed.)
Strategy Cost/Employee, $ Measurement Score
A 5.20 50
B 23.40 182
C 3.75 40
D 10.80 75
E 8.65 53
F 15.10 96
9.47 There are a number of techniques to help people
stop smoking, but their cost and effectiveness vary
widely. One accepted measure of effectiveness of
a program is the percentage of enrollees quitting.
The table below shows several techniques touted
as effective stop-smoking methods, some historical
data on the approximate cost of each program
per person, and the percentage of people smokefree
3 months after the program ended.
Technique
The Cancer Society provides annual cost-offset
funding to cancer patients so more people can afford
these programs. A large clinic in St. Louis has the
capacity to treat each year the number of people
shown. If the clinic plans to place a proposal with the
Cancer Society to treat a specified number of people
annually, estimate the amount of money the clinic
should ask for in its proposal to do the following:
( a) Conduct programs at the capacity level
for the technique with the lowest costeffectiveness
ratio.
(b) Offer programs using as many techniques as
possible to treat up to 1300 people per year
using the most cost-effective techniques.
Cost, $/
Enrollee
% Quitting
Treatment
Capacity per Year
Acupuncture 700 9 250
Subliminal message 150 1 500
Aversion therapy 1700 10 200
Outpatient clinic 2500 39 400
In-patient clinic 1800 41 550
Nicotine replacement
therapy (NRT)
1300 20 100
9.48 The cost in $ per year and effectiveness measure
in items salvaged per year for four mutually
exclusive service sector alternatives have been
collected. ( a ) Calculate the cost effectiveness ratio
for each alternative, and ( b ) use the CER to identify
the best alternative.
Alternative Cost C, $/Year Salvaged Items/Year E
W 355 20
X 208 17
Y 660 41
Z 102 7
9.49 An engineering student has only 30 minutes before
the final exam in statics and dynamics. He wants to

258 Chapter 9 Benefit/Cost Analysis and Public Sector Economics
get help solving a type of problem that he knows
will be on the test from the professor’s review during
the previous class. There is time for using only
one method of assistance before the exam; he must
select well. In a rapid process of estimation, he determines
how many minutes it would take for each
method of assistance and how many points it might
gain for him on the final. The method and estimates
follow. Where should he seek help to be
most effective?
Assistance from
Minutes for Assistance Points Gained
Teaching assistant 15 15
Course slides on Web 20 10
Friend in class 10 5
Professor 20 15
Public Sector Ethics
9.50 During the design and specifications development
stages of a remote meter reading system for residential
electricity use (a system that allows monthly
usage to be transmitted via phone lines with no
need to physically view meters), the two engineers
working on the project for the city of Forest Ridge
noted something different from what they expected.
The first, an electrical/software engineer, noted that
the city liaison staff member provided all the information
on the software options, but only one option,
the one from Lorier Software, was ever discussed
and detailed. The second designer, an
industrial/systems engineer, further noted that all
the hardware specifications provided to them by
this same liaison came from the same distributor,
namely, Delsey Enterprises. Coincidently, at a
weekend family picnic for city employees, to which
the engineers had been invited, they met a couple
named Don Delsey and Susan Lorier. Upon review,
they learned that Don is the son-in-law of the city
liaison and Susan is his stepdaughter. Based on
these observations and before they complete the
system design and specifications, what should the
two engineers do, if anything, about their suspicions
that the city liaison person is trying to bias the
design to favor of the use of his relatives’ software
and hardware businesses?
9.51 Explain the difference between public policy making
and public planning.
9.52 Since transportation via automobile was introduced,
drivers throughout the country of Yalturia in eastern
Europe have driven on the left side of the road. Recently
the Yalturian National Congress passed a law
that within 3 years, a right-hand driving convention
will be adopted and implemented throughout the
country. This is a major policy change for the country
and will require significant public planning and
project development to implement successfully and
safely. Assume you are the lead engineering consultant
from Halcrow Engineers, responsible for developing
and describing many of these major projects.
Identify six of the projects you deem necessary.
State a name and provide a one- or two-sentence
description of each project.
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
9.53 All of the following are examples of public sector
projects, except:
(a) Bridges (b) Emergency relief
(c) Prisons (d) Oil wells
9.54 All of the following are usually associated with
public sector projects except:
(a) Funding from taxes (b) Profit
(c) Disbenefits
(d) Infinite life
9.55 All of the following would be examples of public
projects except:
(a) Air traffic control system
(b) Establishing a dot.com company
(c) Dam with irrigation canals
(d) Mass transit system
9.56 In a conventional B/C ratio, revenue received by
the government from admission to national parks
should be:
(a) Added to benefits in the numerator
(b) Subtracted from benefits in the numerator
(c)
(d)
Subtracted from costs in the denominator
Added to costs in the denominator
9.57 In a conventional B/C ratio:
(a) Disbenefits and M&O costs are subtracted
from benefits.
(b) Disbenefits are subtracted from benefits and
M&O costs are included in costs.
(c) Disbenefits and M&O costs are added to
costs.
(d) Disbenefits are added to costs and M&O
costs are subtracted from benefits.
9.58 In a modified B/C ratio:
( a) Disbenefits are put in the denominator.
( b) Benefits are subtracted from costs.
( c) M&O costs are put in the denominator.
( d) M&O costs are put in the numerator.
9.59 If two mutually exclusive alternatives have B/C
ratios of 1.4 and 1.5 for the lower- and higher-cost
alternatives, respectively, the following is correct:

Case Study 259
( a) The B/C ratio on the increment between
them is equal to 1.5.
( b) The B/C ratio on the increment between
them is between 1.4 and 1.5.
( c) The B/C ratio on the increment between
them is less than 1.5.
( d) The higher-cost alternative is the better one
economically.
9.60 In evaluating three independent alternatives by the B/C
method, the alternatives were ranked A, B, and C,
respectively, in terms of increasing cost, and the following
results were obtained for overall B/C ratios: 1.1,
0.9, and 1.3. On the basis of these results, you should:
( a) Select only alternative A
( b) Select only alternative C
( c) Compare A and C incrementally
( d) Select alternatives A and C
9.61 An alternative has the following cash flows:
Benefits of $50,000 per year
Disbenefits of $27,000 per year
Initial cost of $250,000
M&O costs of $10,000 per year
If the alternative has an infinite life and the interest
rate is 10% per year, the B/C ratio is closest to:
( a) 0.52 (b) 0.66
( c) 0.91 (d) 1.16
9.62 At the interest rate of 10% per year, an alternative
with the following estimates has a modified B/C
ratio that is closest to:
Benefits of $60,000 per year
Disbenefits of $29,000 per year
Amortized first cost of $20,000 per year
M&O costs of $15,000 per year
( a) 0.65 (b) 0.72
( c) 0.80 (d) 1.04
9.63 In evaluating three mutually exclusive alternatives
by the B/C method, the alternatives are ranked A,
B, and C, respectively, in terms of increasing cost,
and the following results are obtained for the
overall B/C ratios : 1.1, 0.9, and 1.06. On the basis
of these results, you should:
( a) Select A
( b) Select C
(c)
(d)
Select A and C
Compare A and C incrementally
9.64 An alternative with an infinite life has a B/C ratio
of 1.5. The alternative has benefits of $50,000 per
year and annual maintenance costs of $10,000 per
year. The first cost of the alternative at an interest
rate of 10% per year is closest to:
(a) $23,300 (b) $85,400
(c) $146,100 (d) $233,000
9.65 Cost-effectiveness analysis (CEA) differs from
cost-benefit (B/C) analysis in that:
(a) CEA cannot handle multiple alternatives.
(b) CEA expresses outcomes in natural units
rather than in currency units.
(c) CEA cannot handle independent alternatives.
(d) CEA is more time-consuming and resourceintensive.
9.66 Several private colleges claim to have programs
that are very effective at teaching enrollees how
to become entrepreneurs. Two programs, identified
as program X and program Y, have produced
4 and 6 persons per year, respectively, who were
recognized as entrepreneurs. If the total cost of
the programs is $25,000 and $33,000, respectively,
the incremental cost-effectiveness ratio is
closest to:
(a) 6250 (b) 5500
(c) 4000 (d) 1333
9.67 The statements contained in a code of ethics are
variously known as all of the following except:
(a) Canons (b) Laws
(c) Standards (d) Norms
9.68 Of the following, the word not related to ethics is:
( a) Virtuous (b) Honest
(c) Lucrative (d) Proper
9.69 All of the following are examples of unethical
behavior except:
(a) Offering services at prices lower than the
competition
(b) Price fixing
(c) Bait and switch
(d) Selling on the black market
CASE STUDY
COMPARING B/C ANALYSIS AND CEA OF TRAFFIC ACCIDENT REDUCTION
Background
This case study compares benefit/cost analysis and costeffectiveness
analysis on the same information about highway
lighting and its role in accident reduction.
Poor highway lighting may be one reason that proportionately
more traffic accidents occur at night. Traffic accidents
are categorized into six types by severity and value.
For example, an accident with a fatality is valued at

260 Chapter 9 Benefit/Cost Analysis and Public Sector Economics
approximately $4 million, while an accident in which there
is property damage (to the car and contents) is valued at
$6000. One method by which the impact of lighting is measured
compares day and night accident rates for lighted and
unlighted highway sections with similar characteristics. Observed
reductions in accidents seemingly caused by too low
lighting can be translated into either monetary estimates of
the benefits B of lighting or used as the effectiveness
measure E of lighting.
Information
Freeway accident data were collected in a 5-year study. The
property damage category is commonly the largest based on
the accident rate. The number of accidents recorded on a section
of highway is presented here.
Number of Accidents Recorded 1
Unlighted
Lighted
Accident Type Day Night Day Night
Property damage 379 199 2069 839
The ratios of night to day accidents involving property damage
for the unlighted and lighted freeway sections are 199/379
0.525 and 839/2069 0.406, respectively. These results indicate
that the lighting was beneficial. To quantify the benefit, the
accident rate ratio from the unlighted section will be applied to
the lighted section. This will yield the number of accidents that
were prevented. Thus, there would have been (2069)(0.525)
1086 accidents instead of 839 if there had not been lights on the
freeway. This is a difference of 247 accidents. At a cost of $6000
per accident, this results in a net annual benefit of
B (247)($6000) $1,482,000
For an effectiveness measure of number of accidents prevented,
this results in E 247.
To determine the cost of the lighting, it will be assumed that
the light poles are center poles 67 meters apart with 2 bulbs
each. The bulb size is 400 watts, and the installation cost is
$3500 per pole. Since these data were collected over 87.8 kilometers
of lighted freeway, the installed cost of the lighting is
(with number of poles rounded off):
Installation cost $3500 ( 87.8 ———
0.067 )
3500(1310)
$4,585,000
There are a total of 87.8/0.0671310 poles, and electricity
costs $0.10 per kWh. Therefore, the annual power cost is
Annual power cost
1310 poles(2 bulbs/pole)(0.4 kilowatt/bulb)
(12 hours/day)(365 days/year)
($0.10/kilowatt-hour)
$459,024 per year
The data were collected over a 5-year period. Therefore, the
annualized cost C at i 6% per year is
Total annual cost $4,585,000( A/P ,6%,5)
459,024
$1,547,503
If a benefit/cost analysis is the basis for a decision on additional
lighting, the B/C ratio is
B/C —————
1,482,000
1,547,503 0.96
Since B/C 1.0, the lighting is not justified. Consideration
of other categories of accidents is necessary to obtain a better
basis for decisions. If a cost-effectiveness analysis (CEA) is
applied, due to a judgment that the monetary estimates for
lighting’s benefit is not accurate, the C/E ratio is
C/E —————
1,547,503 6265
247
This can serve as a base ratio for comparison when an incremental
CEA is performed for additional accident reduction proposals.
These preliminary B/C and C/E analyses prompted the
development of four lighting options:
W) Implement the plan as detailed above; light poles
every 67 meters at a cost of $3500 per pole.
X) Install poles at twice the distance apart (134 meters).
This is estimated to cause the accident prevention
benefit to decrease by 40%.
Y) Install cheaper poles and surrounding safety guards,
plus slightly lowered lumen bulbs (350 watts) at a
cost of $2500 per pole; place the poles 67 meters
apart. This is estimated to reduce the benefit by 25%.
Z) Install cheaper equipment for $2500 per pole with
350-watt lightbulbs and place them 134 meters apart.
This plan is estimated to reduce the accident prevention
measure by 50% from 247 to 124.
Case Study Exercises
Determine if a definitive decision on lighting can be determined
by doing the following:
1. Use a benefi t/cost analysis to compare the four alternatives
to determine if any are economically justified.
2. Use a cost-effectiveness analysis to compare the four
alternatives.
From an understanding viewpoint, consider the following:
3. How many property-damage accidents could be prevented
on the unlighted portion if it were lighted?
4. What would the lighted, night-to-day accident ratio
have to be to make alternative Z economically justified
by the B/C ratio?
5. Discuss the analysis approaches of B/C and C/E. Does
one seem more appropriate in this type of situation than
the other? Why? Can you think of other bases that might
be better for decisions for public projects such as this one?
1 Portion of data reported in Michael Griffin, “Comparison of the Safety of Lighting on Urban Freeways,” Public Roads, vol. 58, pp. 8–15, 1994.

LEARNING STAGE 2: EPILOGUE
Selecting the Basic Analysis Tool
In the previous five chapters, several equivalent evaluation techniques have been discussed. Any
method—PW, AW, FW, ROR, or B/C—can be used to select one alternative from two or more and
obtain the same, correct answer. Only one method is needed to perform the engineering economy
analysis, because any method, correctly performed, will select the same alternative. Yet different information
about an alternative is available with each different method. The selection of a method and
its correct application can be confusing.
Table LS2–1 gives a recommended evaluation method for different situations, if it is not specified
by the instructor in a course or by corporate practice in professional work. The primary criteria for
selecting a method are speed and ease of performing the analysis. Interpretation of the entries in each
column follows.
Evaluation period: Most private sector alternatives (revenue and cost) are compared over their
equal or unequal estimated lives, or over a specific period of time. Public sector projects are
commonly evaluated using the B/C ratio and usually have long lives that may be considered infinite
for economic computation purposes.
Type of alternatives: Private sector alternatives have cash flow estimates that are revenuebased
(includes income and cost estimates) or cost-based (cost estimates only). For cost alternatives,
the revenue cash flow series is assumed to be equal for all alternatives. For public sector
projects, the difference between costs and timing is used to select one alternative over
another. Service sector projects for which benefits are estimated using a nonmonetary effectiveness
measure are usually evaluated with a method such as cost-effectiveness analysis. This
applies to all evaluation periods.
Recommended method: Whether an analysis is performed by hand, calculator, or spreadsheet, the
method(s) recommended in Table LS2–1 will correctly select one alternative from two or more as
TABLE LS2–1
Recommended Method to Compare Mutually Exclusive Alternatives,
Provided the Method Is Not Preselected
Evaluation Period Type of Alternatives Recommended Method
Equal lives of
alternatives
Unequal lives of
alternatives
Series to
Evaluate
Revenue or cost AW or PW Cash flows
Public sector B/C, based on AW or PW Incremental
cash flows
Revenue or cost AW Cash flows
Public sector B/C, based on AW Incremental
cash flows
Study period Revenue or cost AW or PW Updated cash
flows
Public sector B/C, based on AW or PW Updated incremental
cash flows
Long to infinite Revenue or cost AW or PW Cash flows
Public sector B/C, based on AW Incremental
cash flows

262 Chapter 9 Benefit/Cost Analysis and Public Sector Economics
rapidly as possible. Any other method can be applied subsequently to obtain additional information
and, if needed, verification of the selection. For example, if lives are unequal and the
rate of return is needed, it is best to first apply the AW method at the MARR and then determine
the selected alternative’s i * using the same AW relation with i as the unknown.
Series to evaluate: The estimated cash flow series for one alternative and the incremental
series between two alternatives are the only two options for present worth or annual worth
evaluation. For spreadsheet analyses, this means that the NPV or PV functions (for present
worth) or the PMT function (for annual worth) is applied. The word updated is added as a
reminder that a study period analysis requires that cash flow estimates (especially salvage/
market values) be reexamined and updated before the analysis is performed.
Once the evaluation method is selected, a specific procedure must be followed. These procedures
were the primary topics of the last five chapters. Table LS2–2 summarizes the important
TABLE LS2–2 Characteristics of an Economic Analysis of Mutually Exclusive Alternatives Once the
Evaluation Method Is Determined
Evaluation
Method
Equivalence
Relation
Lives of
Alternatives
Time
Period for
Analysis
Series to
Evaluate
Rate of
Return;
Interest Rate
Decision
Guideline:
Select 1
PW Equal Lives Cash flows MARR Numerically
largest PW
PW Unequal LCM Cash flows MARR Numerically
largest PW
Present worth PW Study period Study period Updated MARR
Numerically
CC
Long to
infinite
cash flows
largest PW
Infinity Cash flows MARR Numerically
largest CC
Future worth FW Same as present worth for equal lives, unequal lives,
and study period
Numerically
largest FW
AW Equal or
unequal
Lives Cash flows MARR Numerically
largest AW
Annual worth AW Study period Study period Updated
cash flows
MARR
Numerically
largest AW
AW Long to
infinite
Infinity Cash flows MARR Numerically
largest AW
PW or AW Equal Lives Incremental
cash flows
Find i* Last i *
MARR
PW or AW Unequal LCM of pair Incremental
cash flows
Find i * Last i *
MARR
Rate of return AW Unequal Lives Cash flows Find i * Last i *
MARR
PW or AW Study period Study period Updated
incremental
cash flows
Find i * Last i *
MARR
PW Equal or
unequal
Benefit/cost AW Equal or
unequal
AW or PW Long to
infinite
LCM of pairs
Lives
Infinity
Incremental
cash flows
Incremental
cash flows
Incremental
cash flows
Discount rate
Discount rate
Discount rate
Last B/C
1.0
Last B/C
1.0
Last B/C
1.0
1 Lowest equivalent cost or largest equivalent income.

Learning Stage 2: Epilogue 263
elements of the procedure for each method—PW, AW, ROR, and B/C. FW is included as an extension
of PW. The meaning of the entries in Table LS2–2 follows.
Equivalence relation The basic equation written to perform any analysis is either a PW
or an AW relation. The capitalized cost (CC) relation is a PW relation for infinite life, and
the FW relation is likely determined from the PW equivalent value. Additionally, as we
learned in Chapter 6, AW is simply PW times the A/P factor over the LCM or study
period.
Lives of alternatives and time period for analysis The length of time for an evaluation (the
n value) will always be one of the following: equal lives of the alternatives, LCM of unequal
lives, specified study period, or infinity because the lives are very long.
• PW analysis always requires the LCM of compared alternatives.
• Incremental ROR and B/C methods require the LCM of the two alternatives being compared.
• The AW method allows analysis over the respective alternative lives.
• CC analysis has an infinite time line and uses the relation P Ai.
The one exception is for the incremental ROR method for unequal-life alternatives using an AW
relation for incremental cash fl ows . The LCM of the two alternatives compared must be used.
This is equivalent to using an AW relation for the actual cash fl ows over the respective lives. Both
approaches find the incremental rate of return i *.
Series to evaluate Either the estimated cash flow series or the incremental series is used to
determine the PW value, the AW value, the i * value, or the B/C ratio.
Rate of return (interest rate) The MARR value must be stated to complete the PW, FW,
or AW method. This is also correct for the discount rate for public sector alternatives analyzed
by the B/C ratio. The ROR method requires that the incremental rate be found in order
to select one alternative. It is here that the dilemma of multiple rates appears, if the sign tests
indicate that a unique, real number root does not necessarily exist for a nonconventional
series.
Decision guideline The selection of one alternative is accomplished using the general guideline
in the rightmost column. Always select the alternative with the numerically largest PW,
FW, or AW value. This is correct for both revenue and cost alternatives. The incremental cash
flow methods—ROR and B/C—require that the largest initial cost and incrementally justified
alternative be selected, provided it is justified against an alternative that is itself justified. This
means that the i * exceeds MARR, or the B/C exceeds 1.0.
EXAMPLE LS2–1
Read through the problem statement of the following examples, neglecting the evaluation
method used in the example. Determine which evaluation method is probably the fastest and
easiest to apply. Is this the method used in the example? (a) 8.6, (b) 6.5, (c) 5.8, (d) 5.4.
Solution
Referring to the contents of Table LS2–1, the following methods should be applied first.
(a) Example 8.6 involves four revenue alternatives with equal lives. Use the AW or PW value
at the MARR of 10%. The incremental ROR method was applied in the example.
(b) Example 6.5 requires selection between three public sector alternatives with unequal lives,
one of which is 50 years and another is infinite. The B/C ratio of AW values is the best
choice. This is how the problem was solved.

264 Chapter 9 Benefit/Cost Analysis and Public Sector Economics
(c) Since Example 5.8 involves two cost alternatives with one having a long life, either AW or
PW can be used. Since one life is long, capitalized cost, based on P A/i, is best in this
case. This is the method applied in the example.
(d) Example 5.4 is in the series of progressive examples. It involves 5-year and 10-year cost
alternatives. The AW method is the best to apply in this case. The PW method for the LCM
of 10 years and a study period of 5 years were both presented in the example.

LEARNING STAGE 3
Making Better Decisions
LEARNING STAGE 3
Making Better
Decisions
CHAPTER 10
Project Financing and
Noneconomic
Attributes
CHAPTER 11
Replacement and
Retention Decisions
CHAPTER 12
Independent
Projects with
Budget Limitation
CHAPTER 13
Breakeven and
Payback Analysis
Most of the evaluations in the real world involve more than
a simple economic selection of new assets or projects. The
chapters in this stage introduce information-gathering and
techniques that make decisions better. For example, noneconomic
parameters can be introduced into the project analysis study
through multiple attribute evaluation, and the appropriate MARR
for a corporation or type of alternative can tailor and improve the
economic decision.
The future is certainly not exact. However, techniques such as
replacement/retention studies, breakeven analysis, and payback
analysis help make informed decisions about future uses of
existing assets and systems.
After completing these chapters, you will be able to go beyond
the basic alternative analysis tools of the previous chapters. The
techniques covered in this learning stage take into consideration the
moving targets of change over time.
Important note: If asset depreciation and taxes are to be considered
by an after-tax analysis, Chapters 16 and 17 should be covered
before or in conjunction with these chapters.

CHAPTER 10
Project
Financing and
Noneconomic
Attributes
LEARNING OUTCOMES
Purpose: Explain debt and equity financing, select the appropriate MARR, and consider multiple attributes when
comparing alternatives.
SECTION TOPIC LEARNING OUTCOME
10.1 COC and MARR • Explain the relation between cost of capital and
the MARR; explain why MARR values vary.
10.2 D-E mix and WACC • Understand debt-to-equity mix and calculate the
weighted average cost of capital.
10.3 Cost of debt capital • Estimate the cost of debt capital, considering tax
advantages.
10.4 Cost of equity capital • Estimate the cost of equity capital and describe its
relation to MARR and WACC.
10.5 High D-E mixes • Demonstrate the connection between high D-E
mixes and corporate (or personal) financial risk.
10.6 Multiple attributes • Develop weights for multiple attributes used in
alternative evaluation and selection.
10.7 Additive weights • Apply the weighted attribute method to
alternative evaluations that include noneconomic
attributes.

T
his chapter discusses the different ways to finance a project through debt and
equity sources and explains how the MARR is established. The descriptions here
complement the introductory material of Chapter 1 on the same topics. Some of
the parameters specified earlier are unspecified here, and in future chapters. As a result,
some of the textbook aspects apparent in previous chapters are removed, thus coming closer
to treating the more complex, real-world situations in which professional practice and decision
making occur.
Until now, only one dimension—the economic one—has been the basis for judging the
economic viability of one project, or the selection basis from two or more alternatives. In this
chapter, guidelines and techniques explain the determination and use of multiple (noneconomic)
attributes helpful in selecting between alternatives.
10.1 MARR Relative to the Cost of Capital
The MARR value used in alternative evaluation is one of the most important parameters of
a study. In Chapter 1 the MARR was described relative to the weighted costs of debt and
equity capital. This and the next four sections explain how to establish a MARR under varying
conditions.
To form the basis for a realistic MARR, the types and cost of each source of project financing
should be understood and estimated. There is a strong connection between the costs of debt and
equity capital and the MARR used to evaluate one or more alternatives, whether they are mutually
exclusive or independent. There are several terms and relationships important to the understanding
of project financing and the MARR that is specified to evaluate projects using PW, AW,
FW, or B/C methods. (Reference to Section 1.9 will complement the following material.)
The cost of capital is the weighted average interest rate paid based on the proportion of
i nvestment capital from debt and equity sources .
The MARR is then set relative to the cost of capital. The MARR can be set for one project, a
series of projects, a division of a corporation, or the entire company. MARR values change
over time due to changing circumstances.
When no specific MARR is established, the estimated net cash flows and available capital
establish an inherent MARR. This rate is determined by finding the ROR ( i *) value of the
project cash flows. This rate is utilized as the opportunity cost , which is the ROR of the first
project not funded due to the lack of capital funds.
Cost of capital
MARR
Opportunity cost
Before we discuss cost of capital, we review the two primary sources of capital.
Debt capital represents borrowing from outside the company, with the principal repaid at a
stated interest rate following a specified time schedule. Debt financing includes borrowing via
bonds, loans, and mortgages. The lender does not share in the profits made using the debt
funds, but there is risk in that the borrower could default on part of or all the borrowed funds.
The amount of outstanding debt financing is indicated in the liabilities section of the corporate
balance sheet.
Equity capital is corporate money comprised of the funds of owners and retained earnings.
Owners’ funds are further classified as common and preferred stock proceeds or owners’ capital
for a private (non-stock- issuing) company. Retained earnings are funds previously retained
in the corporation for capital investment. The amount of equity is indicated in the net worth
section of the corporate balance sheet.
To illustrate the relation between cost of capital and MARR, assume a new greenhouse gas
emission control system will be completely financed by a $25,000,000 bond issue (100% debt
financing), and assume the dividend rate on the bonds is 8%. Therefore, the cost of debt capital
is 8% as shown in Figure 10–1 . This 8% is the minimum for MARR. Management may increase
this MARR in increments that reflect its desire for added return and its perception of risk. For
example, management may add an amount for all capital commitments in this area. Suppose this
amount is 2%. This increases the expected return to 10% ( Figure 10–1 ). Also, if the risk associated
with the investment is considered substantial enough to warrant an additional 1% return requirement,
the final MARR is 11%.

268 Chapter 10 Project Financing and Noneconomic Attributes
Established
MARR
11%
Risk factor added
1%
10% Expected return
requirement added
2%
Minimum
MARR
8% Cost of capital
Figure 10–1
A fundamental relation between cost of capital
and MARR used in practice.
The recommended approach does not follow the logic presented above. Rather, the cost of
capital (8% here) should be the established MARR. Then the i * value is determined from the
estimated net cash flows. Using this approach, suppose the control system is estimated to return
11%. Now, additional return requirements and risk factors are considered to determine if 3%
above the MARR of 8% is sufficient to justify the capital investment. After these considerations,
if the project is not funded, the effective MARR is now 11%. This is the opportunity cost discussed
previously—the unfunded project i * has established the effective MARR for emission
control system alternatives at 11%, not 8%.
The setting of the MARR for an economy study is not an exact process. The debt and equity
capital mix changes over time and between projects. Also, the MARR is not a fixed value established
corporatewide. It is altered for different opportunities and types of projects. For example,
a corporation may use a MARR of 10% for evaluating the purchase of assets (equipment, cars)
and a MARR of 20% for expansion investments, such as acquiring smaller companies.
The effective MARR varies from one project to another and through time because of factors
such as the following:
Project risk. Where there is greater risk (perceived or actual) associated with proposed
projects, the tendency is to set a higher MARR. This is encouraged by the higher cost of debt
capital for projects considered risky. This usually means that there is some concern that the
project will not realize its projected revenue requirements.
Investment opportunity. If management is determined to expand in a certain area, the
MARR may be lowered to encourage investment with the hope of recovering lost revenue in
other areas. This common reaction to investment opportunity can create havoc when the
guidelines for setting a MARR are too strictly applied. Flexibility becomes very important.
Government intervention. Depending upon the state of the economy, international relations,
and a host of other factors, the federal government (and possibly lower levels) can dictate
the forces and direction of the free market. This may occur through price limits, subsidies,
import tariffs, and limitation on availability. Both short-term and long-term government interventions
are commonly present in different areas of the economy. Examples are steel imports,
foreign capital investment, car imports, and agricultural product exports. During the time that
such government actions are in force, there is a strong impact to increase or decrease taxes,
prices, etc., thus tending to move the MARR up or down.
Tax structure. If corporate taxes are rising (due to increased profits, capital gains, local
taxes, etc.), pressure to increase the MARR is present. Use of after-tax analysis may assist in
eliminating this reason for a fluctuating MARR, since accompanying business expenses will
tend to decrease taxes and after-tax costs.
Limited capital. As debt and equity capital become limited, the MARR is increased. If the
demand for limited capital exceeds supply, the MARR may tend to be set even higher. The
opportunity cost has a large role in determining the MARR actually used.

10.2 Debt-Equity Mix and Weighted Average Cost of Capital 269
Market rates at other corporations. If the MARR increases at other corporations, especially
competitors, a company may alter its MARR upward in response. These variations are
often based on changes in interest rates for loans, which directly impact the cost of capital.
If the details of after-tax analysis are not of interest, but the effects of income taxes are important,
the MARR may be increased by incorporating an effective tax rate using the formula
Before-tax MARR
after-tax MARR
————————
1 tax rate
[10.1]
The total or effective tax rate, including federal, state, and local taxes, for most corporations is in
the range of 30% to 50%. If an after-tax rate of return of 10% is required and the effective tax rate
is 35%, the MARR for the before-tax economic analysis is 10%/(1 0.35) 15.4%.
EXAMPLE 10.1
Twin brother and sister, Carl and Christy graduated several years ago from college. Carl, an
architect, has worked in home design with Bulte Homes since graduation. Christy, a civil engineer,
works with Butler Industries in structural components and analysis. They both reside in
Richmond, Virginia. They have started a creative e-commerce network through which
Virginia-based builders can buy their “spec home” plans and construction materials much
more cheaply. Carl and Christy want to expand into a regional e-business corporation. They
have gone to the Bank of America (BA) in Richmond for a business development loan. Identify
some factors that might cause the loan rate to vary when BA provides the quote. Also, indicate
any impact on the established MARR when Carl and Christy make economic decisions for
their business.
Solution
In all cases the direction of the loan rate and the MARR will be the same. Using the six factors
mentioned above, some loan rate considerations are as follows:
Project risk: The loan rate may increase if there has been a noticeable downturn in housing
starts, thus reducing the need for the e-commerce connection.
Investment opportunity: The rate could increase if other companies offering similar services
have already applied for a loan at other BA branches regionally or nationwide.
Government intervention: The loan rate may decrease if the federal government has
recently offered Federal Reserve loan money at low rates to banks. The intervention may
be designed to boost the housing economic sector in an effort to offset a significant
slowdown in new home construction.
Taxes: If the state recently removed house construction materials from the list of items
subject to sales tax, the rate might be lowered slightly.
Capital limitation: Assume the computer equipment and software rights held by Carl and
Christy were bought with their own funds and there are no outstanding loans. If additional
equity capital is not available for this expansion, the rate for the loan (debt capital)
should be lowered.
Market loan rates: The local BA branch probably obtains its development loan money
from a large national pool. If market loan rates to this BA branch have increased, the rate
for this loan will likely increase, because money is becoming “tighter.”
10.2 Debt-Equity Mix and Weighted
Average Cost of Capital
The debt-to-equity (D-E) mix identifies the percentages of debt and equity financing for a corporation.
A company with a 40–60 D-E mix has 40% of its capital originating from debt capital
sources (bonds, loans, and mortgages) and 60% derived from equity sources (stocks and retained

270 Chapter 10 Project Financing and Noneconomic Attributes
Figure 10–2
General shape of different
cost of capital curves.
Equity
Cost of capital for each source
Normal
operating
range
WACC
Minimum
WACC
Debt
0% debt
45% 100% debt
Percent debt capital (D-E mix)
earnings). Most projects are funded with a combination of debt and equity capital made available
specifically for the project or taken from a corporate pool of capital. The weighted average cost
of capital (WACC) of the pool is estimated by the relative fractions from debt and equity sources.
If known exactly, these fractions are used to estimate WACC; otherwise the historical fractions
for each source are used in the relation
WACC (equity fraction)(cost of equity capital)
(debt fraction)(cost of debt capital) [10.2]
The two cost terms are expressed as percentage interest rates.
Since virtually all corporations have a mixture of capital sources, the WACC is a value between
the debt and equity costs of capital. If the fraction of each type of equity financing—
common stock, preferred stock, and retained earnings—is known, Equation [10.2] is expanded.
WACC (common stock fraction)(cost of common stock capital)
(preferred stock fraction)(cost of preferred stock capital)
(retained earnings fraction)(cost of retained earnings capital)
(debt fraction)(cost of debt capital) [10.3]
Figure 10–2 indicates the usual shape of cost of capital curves. If 100% of the capital is derived
from equity or 100% is from debt sources, the WACC equals the cost of capital of that source of
funds. There is virtually always a mixture of capital sources involved for any capitalization program.
As an illustration only, Figure 10–2 indicates a minimum WACC at about 45% debt capital.
Most firms operate over a range of D-E mixes. For example, a range of 30% to 50% debt financing
for some companies may be very acceptable to lenders, with no increases in risk or MARR. However,
another company may be considered “risky” with only 20% debt capital. It takes knowledge
about management ability, current projects, and the economic health of the specific industry to
determine a reasonable operating range of the D-E mix for a particular company.
EXAMPLE 10.2
Historically, Hong Kong has imported over 95% of its fresh vegetables each day. In an effort
to develop sustainable and renewable vegetable sources, a new commercial vertical crop technology
is being installed through a public-private partnership with Valcent Products. 1 For
1
“Valcent Announces Agreement to Supply Verticrop Vertical Farming Technology to Hong Kong’s
VF Innovations Ltd.,” www.valcent.net, June 16, 2010 news release.

10.3 Determination of the Cost of Debt Capital 271
illustration purposes, assume that the present worth of the total system cost is $20 million with
financing sources and costs as follows.
Commercial loan for debt financing
Retained earnings from partnering corporations
Sale of stock (common and preferred)
$10 million at 6.8% per year
$4 million at 5.2% per year
$6 million at 5.9% per year
There are three existing international vertical farming projects with capitalization and WACC
values as follows:
Project 1: $5 million with WACC 1 7.9%
Project 2: $30 million with WACC 2 10.2%
Project 3: $7 million with WACC 3 4.8%
Compare the WACC for the Hong Kong (HK) project with the WACC of the existing projects.
Solution
To apply Equation [10.3] to this new project, the fraction of equity (stock and retained earnings)
and debt financing is needed. These are 0.3 for stock ($6 out of $20 million), 0.2 for retained
earnings, and 0.5 for debt ($10 out of $20 million).
WACC HK 0.3(5.9%) 0.2(5.2%) 0.5(6.8%) 6.210%
To correctly weight the other three project WACCs by size, determine the fraction in each one
of the $42 million in total capital: project 1 has $5 million/$42 million 0.119; project 2 has
0.714; project 3 has 0.167. The WACC weighted by project size is WACC W .
WACC W 0.119(7.9%) 0.714(10.2%) 0.167(4.8%) 9.025%
The Hong Kong project has a considerably lower cost of capital than the weighted average of
other projects, considering all sources of funding.
The WACC value can be computed using before-tax or after-tax values for cost of capital. The
after-tax method is the correct one since debt financing has a distinct tax advantage, as discussed
in Section 10.3 below. Approximations of after-tax or before-tax cost of capital are made using
the effective tax rate T e in the relation
After-tax cost of debt capital (before-tax cost)(1 T e ) [10.4]
The effective tax rate is a combination of federal, state, and local tax rates. They are reduced to a
single number T e to simplify computations. Equation [10.4] may be used to approximate the cost
of debt capital separately or inserted into Equation [10.2] for an after-tax WACC rate. Chapter 17
treats taxes and after-tax economic analysis in detail.
10.3 Determination of the Cost of Debt Capital
Debt financing includes borrowing, primarily via bonds and loans. (We learned about bonds in
Section 7.6.) In most industrialized countries, bond dividends and loan interest payments are taxdeductible
as a corporate expense. This reduces the taxable income base upon which taxes are
calculated, with the end result of less taxes paid. The cost of debt capital is, therefore, reduced
because there is an annual tax savings of the expense cash flow times the effective tax rate T e .
This tax savings is subtracted from the debt-capital expense in order to calculate the cost of debt
capital. In formula form,
Tax savings (expenses) (effective tax rate) expenses ( T e ) [10.5]
Net cash flow expenses tax savings expenses (1 T e ) [10.6]
To fi nd the cost of debt capital, develop a PW- or AW-based relation of the net cash flow
(NCF) series with i * as the unknown. Find i * by trial and error, by calculator, or by the RATE

272 Chapter 10 Project Financing and Noneconomic Attributes
or IRR function on a spreadsheet. This is the cost of debt capital used in the WACC computation,
Equation [10.2].
EXAMPLE 10.3
AT&T will generate $5 million in debt capital by issuing five thousand $1000 8% per year
10-year bonds. If the effective tax rate of the company is 30% and the bonds are discounted
2%, compute the cost of debt capital ( a ) before taxes and ( b ) after taxes from the company
perspective. Obtain the answers by hand and spreadsheet.
Solution by Hand
(a) The annual bond dividend is $1000(0.08) $80, and the 2% discounted sales price is $980
now. Using the company perspective, find the i * in the PW relation
0 980 80( P / A , i *,10) 1000( P / F , i *,10)
i * 8.3%
The before-tax cost of debt capital is i * 8.3%, which is slightly higher than the
8% bond interest rate, because of the 2% sales discount.
( b) With the allowance to reduce taxes by deducting the bond dividend, Equation [10.5]
shows a tax savings of $80(0.3) $24 per year. The bond dividend amount for the PW
relation is now $80 24 $56. Solving for i * after taxes reduces the cost of debt
capital to 5.87%.
Solution by Spreadsheet
Figure 10–3 is a spreadsheet image for both before-tax (column B) and after-tax (column C)
analysis using the IRR function. The after-tax net cash flow is calculated using Equation [10.6]
with T e 0.3.
Bond dividend before taxes
1000 * 0.08
Bond dividend after taxes
(1000 * 0.08) * (1 0.3)
IRR(C3:C13)
Figure 10–3
Use of IRR function to determine cost of debt capital before taxes and after taxes, Example10.3.
EXAMPLE 10.4
LST Trading Company will purchase a $20,000 ten-year-life asset. Company managers
have decided to put $10,000 down now from retained earnings and borrow $10,000 at an
interest rate of 6%. The simplified loan repayment plan is $600 in interest each year, with
the entire $10,000 principal paid in year 10. ( a ) What is the after-tax cost of debt capital if
the effective tax rate is 42%? ( b ) How are the interest rate and cost of debt capital used to
calculate WACC?

10.4 Determination of the Cost of Equity Capital and the MARR 273
Solution
( a) The after-tax net cash flow for interest on the $10,000 loan is an annual amount of
600(1 0.42) $348 by Equation [10.6]. The loan repayment is $10,000 in year 10.
PW is used to estimate a cost of debt capital of 3.48%.
0 10,000 348( P / A , i *,10) 10,000( P / F , i *,10)
(b) The 6% annual interest on the $10,000 loan is not the WACC because 6% is paid only on
the borrowed funds. Nor is 3.48% the WACC, since it is only the cost of debt capital. The
cost of the $10,000 equity capital is needed to determine the WACC.
10.4 Determination of the Cost of Equity
Capital and the MARR
Equity capital is usually obtained from the following sources:
Sale of preferred stock
Sale of common stock
Use of retained earnings
Use of owner’s private capital
The cost of each type of financing is estimated separately and entered into the WACC computation.
A summary of one commonly accepted way to estimate each source’s cost of capital is
presented here. One additional method for estimating the cost of equity capital via common stock
is presented. There are no tax savings for equity capital, because dividends paid to stockholders
and owners are not tax-deductible.
Issuance of preferred stock carries with it a commitment to pay a stated dividend annually.
The cost of capital is the stated dividend percentage, for example, 10%, or the dividend amount
divided by the price of the stock. Preferred stock may be sold at a discount to speed the sale, in
which case the actual proceeds from the stock should be used as the denominator. For example,
if a 10% dividend preferred stock with a value of $200 is sold at a 5% discount for $190 per
share, there is a cost of equity capital of ($20/$190) 100% 10.53%.
Estimating the cost of equity capital for common stock is more involved. The dividends paid
are not a true indication of what the stock issue will actually cost in the future. Usually a valuation
of the common stock is used to estimate the cost. If R e is the cost of equity capital (in
decimal form),
first-year dividend
R e ———————— expected dividend growth rate
price of stock
DV 1
——
P g [10.7]
The growth rate g is an estimate of the annual increase in returns that the shareholders receive.
Stated another way, it is the compound growth rate on dividends that the company believes is
required to attract stockholders. For example, assume a multinational corporation plans to raise
capital through its U.S. subsidiary for a new plant in South America by selling $2,500,000 worth
of common stock valued at $20 each. If a 5% or $1 dividend is planned for the first year and an
appreciation of 4% per year is anticipated for future dividends, the cost of capital for this common
stock issue from Equation [10.7] is 9%.
R e ——
20 1 0.04 0.09
The retained earnings and owner’s funds cost of equity capital is usually set equal to the
common stock cost, since it is the shareholders and owners who will realize any returns from
projects in which these funds are invested.

274 Chapter 10 Project Financing and Noneconomic Attributes
Figure 10–4
Expected return on common
stock issue using
CAPM.
R m
R e
Market
security
line
1
Premium
increases
for more
risky securities
R m – R f
Premium
R f
Selected
market
portfolio
0 1.0
EXAMPLE 10.5
Once the cost of capital for all planned equity sources is estimated, the WACC is calculated
using Equation [10.3].
A second method used to estimate the cost of common stock capital is the capital asset pricing
model (CAPM) . Because of the fluctuations in stock prices and the higher return demanded
by some corporations’ stocks compared to others, this valuation technique is commonly applied.
The cost of equity capital from common stock R e , using CAPM, is
R e risk-free return premium above risk-free return
R f ( R m R f ) [10.8]
where volatility of a company’s stock relative to other stocks in the market ( 1.0 is
the norm)
R m return on stocks in a defined market portfolio measured by a prescribed index
R f return on a “safe investment,” which is usually the U.S. Treasury bill rate
The term ( R m R f ) is the premium paid above the safe or risk-free rate. The coefficient
(beta) indicates how the stock is expected to vary compared to a selected portfolio of stocks in
the same general market area, often the Standard and Poor’s 500 stock index. If 1.0, the
stock is less volatile, so the resulting premium can be smaller; when 1.0, larger price movements
are expected, so the premium is increased.
Security is a word that identifies a stock, bond, or any other instrument used to develop capital.
To better understand how CAPM works, consider Figure 10–4 . This is a plot of a market security
line, which is a linear fit by regression analysis to indicate the expected return for different
values. When 0, the risk-free return R f is acceptable (no premium). As increases, the
premium return requirement grows. Beta values are published periodically for most stock-issuing
corporations. Once complete, this estimated cost of common stock equity capital can be included
in the WACC computation in Equation [10.3].
The lead software engineer at SafeSoft, a food industry service corporation, has convinced the
president to develop new software technology for the meat and food safety industry. It is envisioned
that processes for prepared meats can be completed more safely and faster using this
automated control software. A common stock issue is a possibility to raise capital if the cost of
equity capital is below 9%. SafeSoft, which has a historical beta value of 1.09, uses CAPM to
determine the premium of its stock compared to other software corporations. The security
market line indicates that a 5% premium above the risk-free rate is desirable. If U.S. Treasury
bills are paying 2%, estimate the cost of common stock capital.

10.5 Effect of Debt-Equity Mix on Investment Risk 275
Solution
The premium of 5% represents the term R m R f in Equation [10.8].
R e 2.0 1.09(5.0) 7.45%
Since this cost is lower than 9%, SafeSoft should issue common stock to finance this new
venture.
In theory, a correctly performed engineering economy study uses a MARR equal to the cost
of the capital committed to the specific alternatives in the study. Of course, such detail is not
known. For a combination of debt and equity capital, the calculated WACC sets the minimum for
the MARR. The most rational approach is to set MARR between the cost of equity capital and
the corporation’s WACC. The risks associated with an alternative should be treated separately
from the MARR determination, as stated earlier. This supports the guideline that the MARR
should not be arbitrarily increased to account for the various types of risk associated with the
cash flow estimates. Unfortunately, the MARR is often set above the WACC because management
does want to account for risk by increasing the MARR.
EXAMPLE 10.6
The Engineering Products Division of 4M Corporation has two mutually exclusive alternatives
A and B with ROR values of i * A 9.2% and i * B 5.9%. The financing scenario is yet unsettled,
but it will be one of the following: plan 1—use all equity funds, which are currently earning
8% for the corporation; plan 2—use funds from the corporate capital pool which is 25% debt
capital costing 14.5% and the remainder from the same equity funds mentioned above. The
cost of debt capital is currently high because the company has narrowly missed its projected
revenue on common stock for the last two quarters, and banks have increased the borrowing
rate for 4M. Make the economic decision on alternative A versus B under each financing
scenario. The MARR is set by the calculated WACC.
Solution
The capital is available for one of the two mutually exclusive alternatives. For plan 1, 100%
equity, the financing is specifically known, so the cost of equity capital is the MARR, that is,
8%. Only alternative A is acceptable; alternative B is not since the estimated return of 5.9%
does not exceed this MARR.
Under financing plan 2, with a D-E mix of 25–75,
WACC 0.25(14.5) 0.75(8.0) 9.625%
Now, neither alternative is acceptable since both ROR values are less than MARR
WACC 9.625%. The selected alternative should be to do nothing, unless one alternative
absolutely must be selected, in which case noneconomic attributes must be considered.
10.5 Effect of Debt-Equity Mix on Investment Risk
The D-E mix was introduced in Section 10.2. As the proportion of debt capital increases, the
calculated cost of capital decreases due to the tax advantages of debt capital.
The leverage offered by larger debt capital percentages increases the riskiness of projects undertaken
by the company. When large debts are already present, additional financing using debt (or
equity) sources gets more difficult to justify, and the corporation can be placed in a situation
where it owns a smaller and smaller portion of itself. This is sometimes referred to as a highly
leveraged corporation .
Inability to obtain operating and investment capital means increased difficulty for the company
and its projects. Thus, a reasonable balance between debt and equity financing is important for

276 Chapter 10 Project Financing and Noneconomic Attributes
the financial health of a corporation. Example 10.7 illustrates the disadvantages of unbalanced
D-E mixes.
EXAMPLE 10.7
Three auto parts manufacturing companies have the following debt and equity capital amounts
and D-E mixes. Assume all equity capital is in the form of common stock.
Company
Debt
($ in Millions)
Amount of Capital
Equity
($ in Millions) D-E Mix (%–%)
A 10 40 20−80
B 20 20 50−50
C 40 10 80−20
Assume the annual revenue is $15 million for each one and that after interest on debt is considered,
the net incomes are $14.4, $13.4, and $10.0 million, respectively. Compute the return on
common stock for each company, and comment on the return relative to the D-E mixes.
Solution
Divide the net income by the stock (equity) amount to compute the common stock return. In
million dollars,
Return A —— 14.4 0.36 (36%)
40
Return B —— 13.4 0.67 (67%)
20
Return C —— 10.0 1.00 (100%)
10
As expected, the return is by far the largest for highly leveraged C, where only 20% of the
company is in the hands of the ownership. The return is excellent, but the risk associated with
this firm is high compared to A, where the D-E mix is only 20% debt.
The use of large percentages of debt fi nancing greatly increases the risk taken by lenders and
stock owners. Long-term confidence in the corporation diminishes, no matter how large the
short-term return on stock.
The leverage of large D-E mixes does increase the return on equity capital, as shown in previous
examples; but it can also work against the owners and investors. A decrease in asset value
will more negatively affect a highly debt-leveraged company compared to one with small leveraging.
Example 10.8 illustrates this fact.
EXAMPLE 10.8
During the last several years, the U.S. airline industry has had financial problems, in part due
to high fuel costs, fewer customers, security problems, government regulations, aging aircraft,
and union dissatisfaction. As a consequence, the D-E mixes of the so-called traditional companies
(American, United, Delta, and others) have become larger on the debt side than is historically
acceptable. Meanwhile, the D-E mixes of so-called low-cost airlines (Southwest, JetBlue,
and others) have suffered, but not to the same degree. In an effort to reduce costs, assume that
three airlines joined forces to cooperate on a range of services (baggage handling, onboard
food preparation, ticket services, and software development) by forming a new company called
FullServe, Inc. This required $5 billion ($5 B) up-front funding from each airline.
Table 10–1 summarizes the D-E mixes and the total equity capitalization for each airline
after its share of $5 B was removed from available equity funds. The percentage of the $5 B
obtained as debt capital was the same proportion as the debt in the company’s D-E mix. For

10.5 Effect of Debt-Equity Mix on Investment Risk 277
TABLE 10–1 Debt and Equity Statistics, Example 10.8
Airline
Company
Corporate D-E
Mix, %
Amount Borrowed,
$ B
Equity Capital
Available, $ B
National 30–70 1.50 5.0
Global 65–35 3.25 3.7
PanAm 91–9 4.55 6.7
example, National had 30% of its capitalization in debt capital; therefore, 30% of $5 B was
borrowed, and 70% was provided from National’s equity fund.
Unfortunately, after a short time, it was clear that the three-way collaborative effort was a
complete failure, and FullServe was dissolved and its assets were distributed or sold for a total
of $3.0 billion, only 20% of its original value. A total of $1.0 billion in equity capital was returned
to each airline. The commercial banks that provided the original loans then required that
the airlines each pay back the entire borrowed amount now, since FullServe was dissolved and
no profit from the venture could be realized. Assuming the loan and equity amounts are the
same as shown in Table 10–1 , determine the resulting equity capital situation for each airline
after it pays off the loan from its own equity funds. Also, describe one impact on each company
as a result of this expensive and risky venture that failed.
Solution
Determine the level of post-FullServe equity capital using the following relation, in $ billions.
Equity capital pre-FullServe level returned capital loan repayment
National: Equity capital 5.0 1.0 1.50 $4.50
Global: Equity capital 3.7 1.0 3.25 $1.45
PanAm: Equity capital 6.7 1.0 4.55 $3.15
Comparing the equity capital levels ( Table 10–1 ) with the levels above indicates that the
FullServe effort reduced equity amounts by 10% for National, 60% for Global, and 53%
for PanAm. The debt capital to fund the failed FullServe effort has affected National
airlines the least, in large part due to its low D-E mix of 30%–70%. However, Global
and PanAm are in much worse shape financially, and they now must maintain business
with a significantly lower ownership level and much reduced ability to obtain future
capital—debt or equity.
The same principles discussed above for corporations are applicable to individuals. The person
who is highly leveraged has large debts in terms of credit card balances, personal loans, and
house mortgages. As an example, assume two engineers each have a take-home amount of
$40,000 after all income tax, social security, and insurance premiums are deducted from their
annual salaries. Further, assume that the cost of the debt (money borrowed via credit cards and
loans) averages 15% per year and that the total debt is being repaid in equal amounts over
20 years. If Jamal has a total debt of $25,000 and Barry owes $100,000, the remaining amount of
the annual take-home pay may be calculated as follows:
Person
Total
Debt, $
Amount Paid, $ per Year
Cost of
Debt at 15%, $ Repayment of Debt, $
Amount Remaining
from $40,000, $
Jamal 25,000 3,750 1,250 35,000
Barry 100,000 15,000 5,000 20,000
Jamal has 87.5% of his base available while Barry has only 50% available.

278 Chapter 10 Project Financing and Noneconomic Attributes
10.6 Multiple Attribute Analysis: Identification
and Importance of Each Attribute
In Chapter 1 the role and scope of engineering economy in decision making were outlined. The
decision-making process explained in that chapter (Figure 1–1) included the seven steps listed on
the right side of Figure 10–5 . Step 4 is to identify the one or multiple attributes (criteria) upon
which the selection will be based. In all prior evaluations, only one attribute—the economic
one—has been identified and used to select the best alternative. The criterion has been the maximization
of the equivalent value of PW, AW, FW, ROR, B/C ratio, or the CER for service projects.
As we are all aware, most evaluations do and should take into account multiple attributes in
decision making. These are the factors labeled as noneconomic in step 5 of Figure 1–1. However,
these noneconomic dimensions tend to be intangible and often difficult, if not impossible, to
quantify with economic and other scales. Nonetheless, among the many attributes that can be
identified, there are key ones that must be considered in earnest before the alternative selection
process is complete. This section and the next describe some of the techniques that accommodate
multiple attributes in an engineering study.
Multiple attributes enter into the decision-making process in many studies. Public and service sector
projects are excellent examples of multiple-attribute problem solving. For example, the proposal
to construct a dam to form a lake in a low-lying area or to widen the catch basin of a river usually has
several purposes, such as flood control; drinking water; industrial use; commercial development;
recreation; nature conservation for fish, plants, and birds; and possibly other less obvious purposes.
High levels of complexity are introduced into the selection process by the multiple attributes thought
to be important in selecting an alternative for the dam’s location, design, environmental impact, etc.
The left side of Figure 10–5 expands steps 4 and 5 to consider multiple attributes. The discussion
below concentrates on the expanded step 4 and the next section focuses on the evaluation
measure and alternative selection of step 5.
4-1 Attribute Identification Attributes to be considered in the evaluation methodology can
be identified and defined by several methods, some much better than others depending upon the
situation surrounding the study itself. To seek input from individuals other than the analyst is
important; it helps focus the study on key attributes. The following is an incomplete listing of
ways in which key attributes are identified.
• Comparison with similar studies that include multiple attributes
• Input from experts with relevant past experience
• Surveys of constituencies (customers, employees, managers) impacted by the alternatives
• Small group discussions using approaches such as focus groups, brainstorming, or nominal
group technique
• Delphi method, which is a progressive procedure to develop reasoned consensus from different
perspectives and opinions
Figure 10–5
Expansion of the decisionmaking
process to include
multiple attributes.
Consider multiple attributes
Emphasis on one attribute
1. Understand the problem; define the objective.
2. Collect relevant information; define alternatives.
3. Make estimates.
4-1. Identify the attributes for decision
making.
4-2. Determine the relative
importance (weights) of attributes.
4-3. For each alternative, determine
each attribute’s value rating.
4. Identify the selection criteria (one or more
attributes).
5. Evaluate each alternative using
a multiple-attribute technique.
Use sensitivity analysis for key
attributes.
5. Evaluate each alternative; use sensitivity and
risk analysis.
6. Select the best alternative.
7. Implement the solution and monitor results.

10.6 Multiple Attribute Analysis: Identification and Importance of Each Attribute 279
As an illustration, assume that Delta Airlines has decided to purchase five new Boeing 787s
for overseas flights, primarily between the North American west coast and Asian cities, principally
Hong Kong, Tokyo, and Singapore. There are approximately 8000 options for each plane
that must be decided upon by Delta’s engineering, purchasing, maintenance, and marketing personnel
before the order to Boeing is placed. Options range in scope from the material and color
of the plane’s interior to the type of latching devices used on the engine cowlings, and in function
from maximum engine thrust to pilot instrument design. An economic study based on the equivalent
AW of the estimated passenger income per trip has determined that 150 of these options are
clearly advantageous. But other noneconomic attributes are to be considered before some of the
more expensive options are specified. A Delphi study was performed using input from 25 individuals.
Concurrently, option choices for another, unidentified airline’s recent order were shared with
Delta personnel. From these two studies it was determined that there are 10 key attributes for options
selection. Four of the most important attributes are
• Repair time: mean time to repair or replace (MTTR) if the option is or affects a flight-critical
component.
• Safety: mean time to failure (MTTF) of flight-critical components.
• Economic: estimated extra revenue for the option. (Basically, this is the attribute evaluated by
the economic study already performed.)
• Crewmember needs: some measure of the necessity and/or benefits of the option as judged by
representative crewmembers—pilots and attendants.
The economic attribute of extra revenue may be considered an indirect measure of customer satisfaction,
one that is more quantitative than customer opinion/satisfaction survey results. Of course, there
are many other attributes that can be, and are, used. However, the point is that the economic study
may directly address only one or a few of the key attributes vital to alternative decision making.
An attribute routinely identified by individuals and groups is risk .
Risk is a possible variation in a parameter from an expected, desired, or predicted value that
may be detrimental to observing the intended outcome(s) of the product, process, or system.
It represents the absence of or deviation from certainty. Risk is present when there are two or
more observable values of a parameter and it is possible to assume or estimate the chance that
each value may occur.
Risk
Actually, risk is not a stand-alone attribute, because it is a part of every attribute in one form or
another. Considerations of variation, probabilistic estimates, etc., in the decision-making process
are treated in Chapters 18 and 19. Formalized sensitivity analysis, expected values, simulation,
and decision trees are some of the techniques useful in handling risk.
4-2 Importance (Weights) for the Attributes Determination of the extent of importance
for each attribute i results in a weight W i that is incorporated into the final evaluation measure. The
weight, a number between 0 and 1, is based upon the experienced opinion of one individual or a
group of persons familiar with the attributes, and possibly the alternatives. If a group is utilized to
determine the weights, there must be consensus among the members for each weight. Otherwise,
some averaging technique must be applied to arrive at one weight value for each attribute.
Table 10–2 is a tabular layout of attributes and alternatives used to perform a multiple attribute
evaluation. Weights W i for each attribute are entered on the left side. The remainder of the table
is discussed as we proceed through steps 4 and 5 of the expanded decision-making process.
Attribute weights are usually normalized such that their sum over all the alternatives is 1.0. This
normalizing implies that each attribute’s importance score is divided by the sum S over all attributes.
Expressed in formula form, these two properties of weights for attribute i (i 1, 2, . . . , m ) are
m
Normalized weights: W i 1.0 [10.9]
i1
importance score
Weight calculation: W i ————————— i
m
importance score i
i 1
importance score i
————————
S
[10.10]

280 Chapter 10 Project Financing and Noneconomic Attributes
TABLE 10–2 Tabular Layout of Attributes and Alternatives
Used for Multiple Attribute Evaluation
Alternatives
Attributes Weights 1 2 3 . . . n
1 W 1
2 W 2
3 W 3
Value ratings V ij
m
W m
Of the many procedures developed to assign weights to an attribute, an analyst is likely to rely
upon one that is relatively simple, such as equal weighting, rank order, or weighted rank order.
Pairwise comparison is another technique. Each is briefly presented below.
Equal Weighting All attributes are considered to be of approximately the same importance
, or there is no rationale to distinguish the more important from the less important attribute.
This is the default approach. Each weight in Table 10–2 will be 1 m , according to Equation
[10.10]. Alternatively, the normalizing can be omitted, in which case each weight is 1 and
their sum is m . In this case, the final evaluation measure for an alternative will be the sum over
all attributes.
Rank Order The m attributes are ordered (ranked) by increasing importance with a score of 1
assigned to the least important and m assigned to the most important. By Equation [10.10], the
weights follow the pattern 1 S , 2 S , . . . , m S . With this method, the difference in weights between
attributes of increasing importance is constant .
Weighted Rank Order The m attributes are again placed in the order of increasing importance.
However, now differentiation between attributes is possible. The most important attribute
is assigned a score, usually 100, and all other attributes are scored relative to it between 100
and 0. Now, define the score for each attribute as s i , and Equation [10.10] takes the form
s i
m
s i
i1
W i ——
[10.11]
This is a very practical method to determine weights because one or more attributes can
be heavily weighted if they are significantly more important than the remaining ones, and
Equation [10.11] automatically normalizes the weights. For example, suppose the four key
attributes in the previous aircraft purchase example are ordered: safety, repair time, crewmember
needs, and economic. If repair time is only one-half as important as safety, and the
last two attributes are each one-half as important as repair time, the scores and weights are
as follows.
Attribute Score Weights
Safety 100 100/200 0.50
Repair time 50 50/200 0.25
Crewmember needs 25 25/200 0.125
Economic 25 25/200 0.125
Sum of scores and weights 200 1.000
Pairwise Comparison Each attribute is compared to each other attribute in a pairwise fashion
using a rating scale that indicates the importance of one attribute over the other. Assume the

10.6 Multiple Attribute Analysis: Identification and Importance of Each Attribute 281
TABLE 10–3 Pairwise Comparison of Three Attributes to Determine Weights
Attribute i 1 Cost 2 Constructability 3 Environment
Cost — 0 1
Constructability 2 — 1
Environment 1 1 —
Sum of scores 3 1 2
Weight W i 0.500 0.167 0.333
three criteria (attributes) upon which a public works project decision is based are cost, constructability,
and environmental impact . Define the importance comparison scale as follows:
0 if attribute is less important than one compared to
1 if attribute is equally important as one compared to
2 if attribute is more important than one compared to
Set up a table listing attributes across the top and down the side, and perform the pairwise comparison
for each column attribute with each row attribute. Table 10–3 presents a comparison with
importance scores included. The arrow to the right of the table indicates the direction of comparison,
i.e., column with row attribute. For example, cost is judged more important than constructability,
thus a score of 2. The complement score of 0 is placed in the reverse comparison of
constructability with cost. The weights are determined by normalizing the scores using Equation
[10.11], where the sum for each column is s i . For the first attribute, cost i 1.
s 1 3
s i 3 1 2 6
Cost weight W 1 36 0.500
Similarly, the other weights are W 2 16 0.167 and W 3 26 0.333.
There are other attribute weighting techniques, especially for group processes, such as utility
functions, and the Dunn-Rankin procedure. These become increasingly sophisticated, but they are
able to provide an advantage that these simple methods do not afford the analyst: consistency of
ranks and scores between attributes and between individuals. If this consistency is important in that
several decision makers with diverse opinions about attribute importance are involved in a study, a
more sophisticated technique may be warranted. There is substantial literature on this topic.
4-3 Value Rating of Each Alternative by Attribute This is the final step prior to calculating
the evaluation measure. Each alternative j is awarded a value rating V ij for each attribute i .
These are the entries within the cells in Table 10–2 . The ratings are appraisals by decision makers
of how well an alternative will perform as each attribute is considered.
The scale for the value rating can vary depending upon what is easiest to understand for those
who do the valuation. A scale of 0 to 100 can be used for attribute importance scoring. However,
the most popular is a scale of 4 or 5 gradations about the perceived ability of an alternative to
accomplish the intent of the attribute. This is called a Likert scale, which can have descriptions
for the gradations (e.g., very poor, poor, good, very good), or numbers assigned between 0 and
10, or 1 to 1, or 2 to 2. The last two scales can give a negative impact to the evaluation
measure for poor alternatives. An example numerical scale of 0 to 10 is as follows:
If You Value the
Alternative as
Give It a Rating
between the Numbers
Very poor 0–2
Poor 3–5
Good 6–8
Very good 9–10
It is preferable to have a Likert scale with four choices (an even number) so that the central tendency
of “fair” is not overrated.

282 Chapter 10 Project Financing and Noneconomic Attributes
TABLE 10–4 Completed Layout for Four Attributes and Three
Alternatives for Multiple Attribute Evaluation
Alternatives
Attributes Weights 1 2 3
Safety 0.50 6 4 8
Repair 0.25 9 3 1
Crew needs 0.125 5 6 6
Economic 0.125 5 9 7
If we now build upon the aircraft purchase illustration to include value ratings, the cells are
filled with ratings awarded by a decision maker. Table 10–4 includes example ratings V ij and the
weights W i determined above. Initially, there will be one such table for each decision maker. Prior
to calculating the final evaluation measure R j , the ratings can be combined in some fashion; or a
different R j can be calculated using each decision maker’s ratings. Determination of this evaluation
measure is discussed below.
10.7 Evaluation Measure for Multiple Attributes
The need for an evaluation measure that accommodates multiple attributes is indicated in step 5
of Figure 10–5 . The measure can be one that attempts to retain all of the ratings, values, and
complexity of previous assessments by multiple decision makers, or the measure can reduce
these inputs to a single-dimension measure. This section introduces a single-dimension measure
that is widely accepted.
A single-dimension measure effectively combines the different aspects addressed by the attribute
importance weights W i and the alternative value ratings V ij . The resulting evaluation measure
is a formula that calculates an aggregated measure for use in selecting from two or more
alternatives. The approach applied in this process is called the rank-and-rate method .
This reduction process removes much of the complexity of trying to balance the different attributes;
however, it also eliminates much of the robust information captured by the process of ranking attributes
for their importance and rating each alternative’s performance against each attribute.
There are additive, multiplicative, and exponential measures, but by far the most commonly
applied is the additive model. The most used additive model is the weighted attribute method ,
also called the additive weight technique . The evaluation measure, symbolized by R j for each
alternative j , is defined as
R j sum of (weight value rating)
m
W i V ij [10.12]
i1
The W i numbers are the attribute importance weights, and V ij is the value rating by attribute i for
each alternative j . If the attributes are of equal weight (also called unweighted ), all W i 1 m , as
determined by Equation [10.10]. This means that W i can be moved outside of the summation in
the formula for R j . (If an equal weight of W i 1.0 is used for all attributes, in lieu of 1 m , then
the R j value is simply the sum of all ratings for the alternative.)
The selection guideline is as follows:
ME alternative selection
Choose the alternative with the largest R j value. This measure assumes that increasing
weights W i mean more important attributes and increasing ratings V ij mean better performance
of an alternative.
Sensitivity analysis for any score, weight, or value rating is used to determine sensitivity of the decision
to it. The Chapter 18 case study includes an example of multiple attribute sensitivity analysis.

Chapter Summary 283
EXAMPLE 10.9
The Island of Niue in the South Pacific Ocean (www.niueisland.com) released a request for proposal
(RFP) 1 for a new or reconditioned workboat as it upgrades the infrastructure of its port services.
The spreadsheet in Figure 10–6 , left two columns, presents the attributes and normalized
weights W i published in the RFP for use in selecting one of the tenders presenting proposals. Four
acceptable proposals were received. The next four columns (C through F ) include value ratings
between 0 and 100 developed by a group of decision makers when the details of each proposal were
evaluated against each attribute. For example, proposal 2 received a perfect score of 100 on delivery
time, but lifetime costs were considered too high (rating of 20) and the price was considered relatively
high (rating of 55). Use these weights and ratings to determine which proposal to pursue first.
Solution
Assume an additive weighting model is appropriate and apply the weighted attribute method. Equation
[10.12] determines the R j measure for the four alternatives. As an illustration, for proposal 3,
R 3 0.30(95) 0.20(60) 0.05(90) 0.35(85) 0.10(100)
28.5 12.0 4.5 29.8 10.0
84.8
The four totals in Figure 10–6 (columns G through J, row 8) indicate that proposal 3 is the
overall best choice for the attributes and weights published in the RFP.
Comment
Any economic measure can be incorporated into a multiple attribute evaluation using this method.
All measures of worth—PW, AW, ROR, B/C, and C/E—can be included; however, their impact
on the final selection will vary relative to the importance placed on the noneconomic attributes.
Figure 10–6
Attributes, weights, ratings, and evaluation measure for Niue workboat proposals, Example 10.9.
1 Used with permission of Government of Niue, Infrastructure Department, “Request for Proposal: Supply
of Workboat,” released March 30, 2010, www.gov.nu/Documents/workboattender3022.pdf.
CHAPTER SUMMARY
The interest rate at which the MARR is established depends principally upon the cost of capital
and the mix between debt and equity financing. The MARR is strongly influenced by the weighted
average cost of capital (WACC). Risk, profit, and other factors can be considered after the AW,
PW, or ROR analysis is completed and prior to final alternative selection. A high debt-to-equity
mix can significantly increase the riskiness of a project and make further debt financing difficult
to acquire for the corporation.
If multiple attributes, which include more than the economic dimension of a study, are to be
considered in making the alternative decision, first the attributes must be identified and their
relative importance assessed. Then each alternative can be value-rated for each attribute. The
evaluation measure is determined using a model such as the weighted attribute method, where the
measure is calculated by Equation [10.12]. The largest value indicates the best alternative.

284 Chapter 10 Project Financing and Noneconomic Attributes
PROBLEMS
Working with MARR
10.1 List at least three factors that affect the MARR,
and discuss how each one affects it.
10.2 State whether each of the following involves debt
financing or equity financing.
(a) $10,000 taken from one partner’s savings account
to pay for equipment repair
(b) Issuance of preferred stock worth $1.3 million
(c) Short-term loan of $75,000 from a local bank
(d) Issuance of $3 million worth of 20-year
bonds
(e) Del Engineering buyback of $8 million of its
own stock using internal funds
10.3 Helical Products, Inc. uses an after-tax MARR of
12% per year. If the company’s effective tax rate
(federal, state, and local taxes) is 40%, determine
the company’s before-tax MARR.
10.4 The owner of a small pipeline construction company
is trying to figure out how much he should
bid in his attempt to win his first “big” contract. He
estimates that his cost to complete the project will
be $7.2 million. He wants to bid an amount that
will give him an after-tax rate of return of 15% per
year if he gets the job, but he doesn’t know how
much he should bid on a before-tax basis. He told
you that his effective state tax rate is 7% and his
effective federal tax rate is 22%.
(a) The expression for determining the overall
effective tax rate is
state rate (1 state rate)(federal rate)
What should his before-tax MARR be in
order for him to make an after-tax MARR
of 15% per year?
(b) How much should he bid on the job?
10.5 A group of private equity investors provided
$16 million to a start-up company involved in
making high-technology detection systems for
drugs and other types of contraband. Immediately
after the investment was made, another investment
opportunity came up for which the investors
didn’t have enough capital. That project would
have yielded an estimated rate of return of 29%
per year before taxes. If the group’s effective tax
rate is 32%, what after-tax rate of return would
the forgone project have yielded?
10.6 Five projects were ranked in decreasing order by
two measures—rate of return (ROR) and present
worth (PW)—to determine which ones should be
Project
ROR,
%
funded, with the total initial investment not to
exceed $18 million. Use the results below to determine
the opportunity cost for each measure.
PW at Initial
15%, Investment,
$1000 $1000 Project
Ranking by ROR
Ranking by PW
Cumulative
Cumulative
Investment, Investment,
$1000 Project $1000
A 44.5 7,138 8,000 A 8,000 A 8,000
B 12.8 1,162 15,000 E 13,000 C 16,000
C 20.4 1,051 8,000 C 21,000 E 21,000
D 9.6 863 8,000 B 36,000 D 29,000
E 26.0 936 5,000 D 44,000 B 44,000
10.7 Tom, the owner of Burger Palace, determined that
his weighted average cost of capital is 8%. He expects
a return of 4% per year on all of his investments.
A proposal presented by the owner of the
Dairy Choice next door seems quite risky to Tom,
but is an intriguing partnership opportunity. Tom
has determined that the proposal’s “risk factor”
will require an additional 3% per year return for
him to accept it.
(a) Use the recommended approach to determine
the MARR that Tom should use, and explain
how the 3% risk factor is compensated for in
this MARR.
(b) Determine the effective MARR for his business
if Tom turns down the proposal.
D-E Mix and WACC
10.8 Electrical generators produce not only electricity,
but also heat from conductor resistance and from
friction losses in bearings. A company that manufactures
generator coolers for nuclear and gas turbine
power plants undertook a plant expansion
through financing that had a debt-equity mix of
45–55. If $18 million came from mortgages and
bond sales, what was the total amount of the
financing?
10.9 Determine the debt-to-equity mix when Applied
Technology bought out Southwest Semiconductor
using financing as follows: $12 million from mortgages,
$5 million from retained earnings, $10 million
from cash on hand, and $20 million from
bonds.
10.10 Business and engineering seniors are comparing
methods of financing their college education during
their senior year. The business student has
$30,000 in student loans that come due at graduation.
Interest is an effective 4% per year. The
engineering senior owes $50,000, 50% from his
parents with no interest due and 50% from a

Problems 285
credit union loan. This latter amount is also due at
graduation with an effective rate of 7% per year.
( a) What is the D-E mix for each student?
( b) If their grandparents pay the loans in full at
graduation, what are the amounts on the
checks they write for each graduate?
( c) When grandparents pay the full amount at
graduation, what percent of the principal
does the interest represent?
10.11 Two public corporations, First Engineering and
Midwest Development, each show capitalization
of $175 million in their annual reports. The balance
sheet for First Engineering indicates a total
debt of $87 million, and that of Midwest Development
indicates a net worth of $62 million. Determine
the D-E mix for each company.
10.12 Forest Products, Inc. invested $50 million. The
company’s overall D-E mix is 60–40.What is the
return on the company’s equity, if the net income is
$5 million on a revenue base of $6 million?
10.13 Determine the weighted average cost of capital for
a company that manufactures miniature triaxial accelerometers
for space-restricted applications. The
financing profile, with interest rates, is as follows:
$3 million in stock sales at 15% per year, $4 million
in bonds at 9%, and $6 million in retained
earnings at 7% per year.
10.14 Growth Transgenics Enterprises (GTE) is contemplating
the purchase of its rival. One of GTE’s genetics
engineers got interested in the financing
strategy of the buyout. He learned there are two
plans being considered. Plan 1 requires 50% equity
funds from GTE’s retained earnings that currently
earn 9% per year, with the balance borrowed
externally at 6%, based on the company’s excellent
stock rating. Plan 2 requires only 20% equity
funds with the balance borrowed at a higher rate of
8% per year.
( a) Which plan has the lower average cost of
capital?
( b) If GTE’s owners decide that the current corporate
WACC of 8.2% will not be exceeded,
what is the maximum cost of debt capital allowed
for each plan? Are these rates higher
or lower than the current estimates?
10.15 Midac Corporation wants to arrange for $50 million
in capital to finance the manufacturing of a
new consumer product. The current plan is 60%
equity capital and 40% debt financing. Calculate
the WACC for the following scenario:
Equity capital : 60%, or $35 million, via common
stock sales for 40% of this amount that will pay
dividends at a rate of 5% per year, and the remaining
60% from retained earnings, which currently
earn 9% per year.
Debt capital : 40%, or $15 million, obtained
through two sources—bank loans for $10 million
borrowed at 8% per year, and the remainder in
convertible bonds at an estimated 10% per year
bond dividend rate.
10.16 A public corporation in which you own common
stock reported a WACC of 11.1% for the year in its
report to stockholders. The common stock that you
own has averaged a total return of 7% per year
over the last 3 years. The annual report also mentions
that projects within the corporation are 75%
funded by its own capital. Estimate the company’s
cost of debt capital.
10.17 BASF will invest $14 million this year to upgrade
its ethylene glycol processes. This chemical is
used to produce polyester resins to manufacture
products varying from construction materials to
aircraft, and from luggage to home appliances. Equity
capital costs 14.5% per year and will supply
65% of the capital funds. Debt capital costs 10%
per year before taxes. The effective tax rate for
BASF is 36%.
(a) Determine the amount of annual revenue
after taxes that is consumed in covering the
interest on the project’s initial cost.
(b) If the corporation does not want to use 65%
of its own funds, the financing plan may include
75% debt capital. Determine the
amount of annual revenue needed to cover
the interest with this plan, and explain the effect
it may have on the corporation’s ability
to borrow in the future.
10.18 A couple planning for their child’s college education
can fund part of or all the expected $100,000
tuition cost from their own funds (through an education
IRA) or borrow all or part of it. The average
return for their own funds is 7% per year, but the
loan is expected to have a higher interest rate as
the loan amount increases. Use a spreadsheet to
generate a plot of the WACC curve with the estimated
loan interest rates below and determine the
best D-E mix for the couple.
Loan Amount, $
Interest Rate, % per year
10,000 5.0
30,000 6.0
50,000 8.0
60,000 9.0
75,000 11.0
100,000 13.0

286 Chapter 10 Project Financing and Noneconomic Attributes
10.19 Over the last few years, Carol’s Fashion Store, a
statewide franchise, has experienced the D-E
mixes and costs of debt and equity capital on several
projects summarized below.
(a) Plot debt, equity, and weighted average cost
of capital.
(b) Determine what mix of debt and equity capital
provided the lowest WACC.
Debt capital
Equity capital
Project Percent Rate Percent Rate
A 100 % 14.5%
B 70 13.0 30% 7.8%
C 65 12.0 35 7.8
D 50 11.5 50 7.9
E 35 9.9 65 9.8
F 20 12.4 80 12.5
G 100 12.5
10.20 For Problem 10.19, use a spreadsheet to ( a ) determine
the best D-E mix and ( b ) determine the best
D-E mix if the cost of debt capital increases by
10% per year, for example, 13.0% increases
to 14.3%.
Cost of Debt Capital
10.21 The cash flow plan associated with a debt financing
transaction allowed a company to receive
$2,800,000 now in lieu of future interest payments
of $196,000 per year for 10 years plus a lump sum
of $2,800,000 in year 10. If the company’s effective
tax rate is 33%, determine the company’s cost
of debt capital ( a ) before taxes and ( b ) after taxes.
10.22 A company that makes several different types of
skateboards, Jennings Outdoors, incurred interest
expenses of $1,200,000 per year from various
types of debt financing. The company borrowed
$19,000,000 in year 0 and repaid the principal of
the loans in year 15 in a lump-sum payment of
$20,000,000. If the company’s effective tax rate is
29%, what was the company’s cost of debt capital
( a ) before taxes and ( b ) after taxes?
10.23 Molex Inc., a manufacturer of cable assemblies for
polycrystalline photovoltaic solar modules, requires
$3.1 million in debt capital. The company
plans to sell 15-year bonds that carry a dividend of
6% per year, payable semiannually. Molex has an
effective tax rate of 32% per year. Determine
( a ) the nominal annual after-tax cost of debt capital
and ( b ) the effective annual after-tax cost of
debt capital.
10.24 Tri-States Gas Producers expects to borrow
$800,000 for field engineering improvements. Two
methods of debt financing are possible—borrow it
all from a bank or issue debenture bonds. The company
will pay an effective 8% per year to the bank
for 8 years. The principal on the loan will be reduced
uniformly over the 8 years, with the remainder of
each annual payment going toward interest. The
bond issue will be for 800 10-year bonds of $1000
each that require a 6% per year dividend payment.
( a) Which method of financing is cheaper after
an effective tax rate of 40% is considered?
( b) Which is the cheaper method using a beforetax
analysis?
10.25 The Sullivan Family Partnership plans to purchase
a refurbished condo in their hometown for investment
purposes. The negotiated $200,000 purchase
price will be financed with 20% of savings (retained
earnings) which consistently makes 6.5%
per year after all relevant income taxes are paid.
Eighty percent will be borrowed at a before-tax
rate of 9% per year for 15 years with the principal
repaid in equal annual installments. If the effective
tax rate is 22% per year, based only on these data,
answer the following.
(a) What is the partnership’s annual loan payment
for each of the 15 years?
(b) What is the net present worth difference between
the $200,000 now and the PW of the
cost of the 80–20 D-E mix series of cash
flows necessary to finance the purchase?
What does this PW value mean?
( c) What is the after-tax WACC for this
purchase?
Cost of Equity Capital
10.26 Determine the cost of equity capital to Hy-Lok
USA if the company sells 500,000 shares of its
preferred stock at a 5% discount from its price of
$130. The stock carries a $10 per year dividend.
10.27 The initial public offering price for the common
stock of SW Refining is $23 per share, and it will
pay a first-year dividend of $0.92 per share. If the
appreciation rate in dividends is anticipated to be
3.2% per year, determine the cost of equity capital
for the stock offering.
10.28 The cost of debt capital is lower after taxes than
before taxes. The cost of equity capital is more difficult
to estimate using the dividend method or the
CAPM model, for example, yet the after-tax and
before-tax cost of equity capital is the same. Why
are the after-tax rates not the same for both types
of financing?
10.29 H2W Technologies is considering raising capital
to expand its offerings of 2-phase and 4-phase

Problems 287
linear stepper motors. The beta value for its stock
is 1.41. Use the capital asset pricing model and a
3.8% premium above the risk-free return to determine
the cost of equity capital if the risk-free return
is 3.2%.
10.30 Management at Hirschman Engineering has asked
you to determine the cost of equity capital based
on the company’s common stock. The company
wants you to use two methods: the dividend
method and the CAPM. Last year, the first year for
dividends, the stock paid $0.75 per share on the
average of $11.50 on the New York Stock Exchange.
Management hopes to grow the dividend
rate at 3% per year. Hirschman Engineering stock
has a volatility that is higher than the norm at 1.3.
If safe investments are returning 5.5% and the 3%
growth on common stocks is also the premium
above the risk-free investments that Hirschman
Engineering plans to pay, calculate the cost of equity
capital using the two methods.
10.31 Common stocks issued by Meggitt Sensing Systems
paid stockholders $0.93 per share on an average
price of $18.80 last year. The company expects
to grow the dividend rate at a maximum of 1.5%
per year. The stock volatility is 1.19, and other
stocks in the same industry are paying an average
of 4.95% per year dividend. U.S Treasury bills are
returning 4.5%. Determine the company’s cost of
equity capital last year using ( a ) the dividend
method and ( b ) the CAPM.
10.32 Last year a Japanese engineering materials corporation,
Yamachi Inc., purchased some U.S. Treasury
bonds that return an average of 4% per year.
Now, Euro bonds are being purchased with a realized
average return of 3.9% per year. The volatility
factor of Yamachi stock last year was 1.10 and has
increased this year to 1.18. Other publicly traded
stocks in this same business are paying an average
of 5.1% dividends per year. Determine the cost of
equity capital for each year, and explain why the
increase or decrease seems to have occurred.
10.33 The engineering manager at FXO Plastics wants to
complete an alternative evaluation study. She
asked the finance manager for the corporate
MARR. The finance manager gave her some data
on the project and stated that all projects must
clear their average (pooled) cost by at least 4%.
Funds Source Amount, $ Average Cost, %
Retained earnings 4,000,000 7.4
Stock sales 6,000,000 4.8
Long-term loans 5,000,000 9.8
(a) Use the data to determine the minimum
MARR.
(b) The study is after taxes and part ( a ) provided
the before-tax MARR. Determine the correct
MARR to use if the tax rate was 32% last year
and the finance manager meant that the 4%
above the cost is for after-tax evaluations.
Different D-E Mixes
10.34 Why is it financially unhealthy for an individual to
maintain a large percentage of debt financing over
a long time, that is, to be highly leveraged?
10.35 In a leveraged buyout of one company by another,
the purchasing company usually obtains borrowed
money and inserts as little of its own equity as possible
into the purchase. Explain some circumstances
under which such a buyout may put the
purchasing company at economic risk.
10.36 Grainger and Company has an opportunity to invest
$500,000 in a new line of direct-drive rotary
screw compressors. Financing will be equally split
between common stock ($250,000) and a loan
with an 8% after-tax interest rate. The estimated
annual NCF after taxes is $48,000 for the next
7 years. The effective tax rate is 50%. Grainger
uses the capital asset pricing model for evaluation
of its common stock. Recent analysis shows that it
has a volatility rating of 0.95 and is paying a premium
of 5% above a safe return on its common
stock. Nationally, the safest investment is currently
paying 3% per year. Is the investment financially
attractive if Grainger uses as the MARR its
( a ) equity cost of capital and ( b ) WACC?
10.37 Fairmont Industries primarily relies on 100% equity
financing to fund projects. A good opportunity
is available that will require $250,000 in capital.
The Fairmont owner can supply the money from
personal investments that currently earn an average
of 7.5% per year. The annual net cash flow
from the project is estimated at $30,000 for the
next 15 years. Alternatively, 60% of the required
amount can be borrowed for 15 years at 7% per
year. If the MARR is the WACC, determine which
plan, if either, should be undertaken. This is a
before-tax analysis.
10.38 Omega Engineering Inc. has an opportunity to invest
$10,000,000 in a new engineering remote
control system for offshore drilling platforms. Financing
will be split between common stock sales
($5,000,000) and a loan with an 8% per year interest
rate. Omega’s share of the annual net cash flow
is estimated to be $1.35 million for each of the
next 5 years. Omega is about to initiate CAPM as

288 Chapter 10 Project Financing and Noneconomic Attributes
its common stock evaluation model. Recent analysis
shows that it has a volatility rating of 1.22 and
is paying a premium of 5% on its common stock
dividend. The U.S. Treasury bills are currently
paying 4% per year. Is the venture financially attractive
if the MARR equals ( a ) the cost of equity
capital and ( b ) the WACC?
10.39 A new annular die process is to be installed for extruding
pipes, tubes, and tubular films. The phase I
installed price for the dies and machinery is
$2,000,000. The manufacturer has not decided
how to finance the system. The WACC over the
last 5 years has averaged 10% per year.
(a)
Two financing alternatives have been defined.
The first requires an investment of
40% equity funds at 9% and a loan for the
balance at an interest rate of 10% per year.
The second alternative requires only 25% equity
funds and the balance borrowed at
10.5% per year. Which approach will result
in the smaller average cost of capital?
(b) Yesterday, the corporate finance committee
decided that the WACC for all new projects
must not exceed the 5-year historical average
of 10% per year. With this restriction, what is
the maximum loan interest rate that can be incurred
for each of the financing alternatives?
10.40 Shadowland, a manufacturer of air-freightable pet
crates, has identified two projects that, though having
a relatively high risk, are expected to move the
company into new revenue markets. Utilize a
spreadsheet solution to ( a ) select any combination
of the projects if the MARR is equal to the aftertax
WACC and ( b ) determine if the same projects
should be selected if the risk factors are enough to
require an additional 2% per year for the investment
to be made.
Project
Initial
Investment, $
After-Tax Cash
Flow, $/Year
Life, Years
Wildlife (W) 250,000 48,000 10
Reptiles (R) 125,000 30,000 5
Financing will be developed using a D-E mix of
60–40 with equity funds costing 7.5% per year.
Debt financing will be developed from $10,000,
5% per year, paid quarterly, 10-year bonds. The effective
tax rate is 30% per year.
10.41 Two friends each invested $20,000 of their own
(equity) funds. Stan, being more conservative, purchased
utility and manufacturing corporation
stocks. Theresa, being a risk taker, leveraged the
$20,000 and purchased a $100,000 condo for rental
property. Considering no taxes, dividends, or revenues,
analyze these two purchases by doing the following
for one year after the funds were invested.
(a) Determine the year-end values of their equity
funds if there was a 10% increase in the
value of the stocks and the condo.
(b) Determine the year-end values of their equity
funds if there was a 10% decrease in the
value of the stocks and the condo.
(c) Use your results to explain why leverage can
be financially risky.
Multiple, Noneconomic Attributes
10.42 In multiple attribute analysis, if three different alternatives
are to be evaluated on the basis of eight
attributes that are considered of equal importance,
what is the weight of each attribute?
10.43 A consulting engineer asked a company manager to
assign importance values (0 to 100) to five attributes
that will be included in an alternative evaluation
process. Determine the weight of each attribute
using the importance scores.
Attribute
Importance Score
1. Safety 60
2. Cost 40
3. Impact 80
4. Environmental 30
5. Acceptability 20
10.44 Ten attributes were rank-ordered in terms of increasing
importance and were identified as A, B,
C, . . . , and J. Determine the weight of ( a ) attribute
C and ( b ) attribute J.
10.45 A team of three people submitted the following
statements about the attributes to be used in a
weighted attribute evaluation. Use the statements
to determine the normalized weights if assigned
scores are between 0 and 100.
Attribute
Comment
1. Flexibility (F) The most important factor
2. Safety (S) 70% as important as uptime
3. Uptime (U) One-half as important as flexibility
4. Rate of return (R) Twice as important as safety
10.46 Different types and capacities of crawler hoes are
being considered for use in a major excavation on
a pipe-laying project. Several supervisors who
served on similar projects in the past have identified
some of the attributes and their view of relative
importance. For the information that follows,
determine the weighted rank order, using a 0-to-10
scale and the normalized weights.

Additional Problems and FE Exam Review Questions 289
Attribute
Comment
1. Truck versus hoe height 90% as important as trenching speed
2. Type of topsoil Only 10% of most important
attribute
3. Type of subsoil 30% as important as trenching speed
4. Hoe cycle time Twice as important as type of
subsoil
5. Hoe trenching speed Most important attribute
6. Pipe laying speed 80% as important as hoe cycle time
10.47 John, who works at Swatch, has decided to use the
weighted attribute method to compare three systems
for manufacturing a watchband. The vice
president and her assistant VP have evaluated each
of three attributes in terms of importance to them,
and John has placed an evaluation from 0 to 100
on each alternative for the three attributes. John’s
ratings for each alternative are as follows:
Alternative
Attribute 1 2 3
Economic return MARR 50 70 100
High throughput 100 60 30
Low scrap rate 100 40 50
Use the weights below to evaluate the alternatives.
Are the results the same for both persons’ weights?
Why?
Importance Score VP Assistant VP
Economic return MARR 20 100
High throughput 80 80
Low scrap rate 100 20
10.48 The Athlete’s Shop has evaluated two proposals
for weight lifting and exercise equipment. A present
worth analysis at i 15% per year of estimated
revenues and costs resulted in PW A
$440,000 and PW B $390,000. In addition to
this economic measure, three more attributes
were independently assigned a relative importance
score from 0 to 100 by the shop manager
and the lead trainer.
Importance Score
Attribute Manager Trainer
Economics 80 80
Durability 35 80
Flexibility 30 100
Maintainability 20 50
Separately, you have used the four attributes to
rate the two equipment proposals on a scale of 0 to
1.0 as shown in the following table. The economic
attribute was rated using the PW values.
Attribute Proposal A Proposal B
Economics 1.00 0.90
Durability 0.35 1.00
Flexibility 1.00 0.90
Maintainability 0.25 1.00
Select the better proposal using each of the following
methods.
(a) Present worth
(b) Weighted evaluations of the shop manager
(c) Weighted evaluations of the lead trainer
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
10.49 The term opportunity cost refers to:
( a) The first cost of an alternative that has been
accepted for funding
( b) The total cost of an alternative that has been
accepted for funding
( c) The rate of return or profit available on the
next-best alternative that had to be forgone
due to lack of capital funds
( d) The cost of an alternative that was not recognized
as an alternative that actually represented
a good opportunity
10.50 The cost of capital is established on the basis of:
( a) The cost of debt financing
( b) The weighted average of debt and equity
financing
(c) The cost of equity financing
(d) The cost of debt financing plus the expected
inflation rate
10.51 All of the following are examples of debt capital
except:
(a) Retained earnings
(b) Long-term bonds
(c) Loan from a local bank
(d) Purchase of equipment using a credit card
10.52 All of the following are examples of equity capital
except:
(a) Sale of preferred stock
(b) Long-term bonds
(c) Company cash on hand
(d) Use of retained earnings

290 Chapter 10 Project Financing and Noneconomic Attributes
10.53 If a public utility expands its capacity to generate
electricity by obtaining $41 million from retained
earnings and $30 million from municipal bond
sales, the utilities’ debt-to-equity mix is closest to:
(a) 58% debt and 42% equity
(b) 73% debt and 27% equity
(c) 27% debt and 73% equity
(d) 42% debt and 58% equity
10.54 Gentech, Inc. financed a new product as follows:
$5 million in stock sales at 13.7% per year, $2 million
in retained earnings at 8.9% per year, and
$3 million through convertible bonds at 7.8% per
year. The company’s WACC is closest to:
(a) 9% per year
(b) 10% per year
(c) 11% per year
(d) 12% per year
10.55 If the after-tax rate of return for a cash flow series
is 11.2% and the corporate effective tax rate is
39%, the approximated before-tax rate of return is
closest to:
(a) 6.8%
(b) 5.4%
(c) 18.4%
(d) 28.7%
10.56 Medzyme Pharmaceuticals has maintained a 50-50
D-E mix for capital investments. Equity capital
has cost 11%; however, debt capital that historically
cost 9% has now increased by 20% per year.
If Medzyme does not want to exceed its historical
weighted average cost of capital (WACC), and it is
forced to go to a D-E mix of 75–25, the maximum
acceptable cost of equity capital is closest to:
(a) 7.6%
(b) 9.2%
(c) 9.8%
(d) 10.9%
10.57 The importance values (0 to 100) for five attributes
are shown below. The weight to assign to attribute
1 is:
(a) 0.16
(b) 0.20
(c) 0.22
(d) 0.25
Attribute
Importance Score
1 55
2 45
3 85
4 30
5 60
10.58 For eight attributes rank-ordered in terms of increasing
importance, the weighting of the sixth attribute
is closest to:
(a) 0.17
(b) 0.14
(c) 0.08
(d) 0.03
CASE STUDY
WHICH IS BETTER—DEBT OR EQUITY FINANCING?
Background
Information
Pizza Hut Corporation has decided to enter the catering business
in three states within its Southeastern U.S. Division,
using the name Pizza Hut At-Your-Place. To deliver the meals
and serving personnel, it is about to purchase 200 vans with
custom interiors for a total of $1.5 million. Each van is expected
to be used for 10 years and have a $1000 salvage
value.
A feasibility study completed last year indicated that the
At-Your-Place business venture could realize an estimated
annual net cash flow of $300,000 before taxes in the three
states. After-tax considerations would have to take into account
the effective tax rate of 35% paid by Pizza Hut.
An engineer with Pizza Hut’s Distribution Division has
worked with the corporate finance office to determine how to
best develop the $1.5 million capital needed for the purchase
of vans. There are two viable financing plans.
Plan A is debt financing for 50% of the capital ($750,000)
with the 8% per year compound interest loan repaid over
10 years with uniform year-end payments. (A simplifying assumption
that $75,000 of the principal is repaid with each
annual payment can be made.)
Plan B is 100% equity capital raised from the sale of $15
per share common stock. The financial manager informed the
engineer that stock is paying $0.50 per share in dividends and
that this dividend rate has been increasing at an average of
5% each year. This dividend pattern is expected to continue,
based on the current financial environment.
Case Study Exercises
1. What values of MARR should the engineer use to determine
the better financing plan?

Case Study 291
2. The engineer must make a recommendation on the financing
plan by the end of the day. He does not know
how to consider all the tax angles for the debt financing
in plan A. However, he does have a handbook that gives
these relations for equity and debt capital about taxes
and cash flows:
Equity capital: no income tax advantages
After-tax net cash flow
(before-tax net cash flow)(1 tax rate)
Debt capital: income tax advantage comes from interest
paid on loans
Taxes (taxable income)(tax rate)
Taxable income net cash flow
loan interest
He decides to forget any other tax consequences and use
this information to prepare a recommendation. Is A or B
the better plan?
3. The division manager would like to know how much
the WACC varies for different D-E mixes, especially
about 15% to 20% on either side of the 50% debt financing
option in plan A. Plot the WACC curve and compare
its shape with that of Figure 10–2 .
After-tax net cash flow before-tax net cash flow
loan principal
loan interest taxes

CHAPTER 11
Replacement
and Retention
Decisions
LEARNING OUTCOMES
Purpose: Perform a replacementretention study between an in-place asset, process, or system and one that could
replace it.
SECTION TOPIC LEARNING OUTCOME
11.1 Replacement study basics • Explain the fundamental approach and
terminology of replacement analysis.
11.2 Economic service life • Determine the ESL that minimizes the total AW
for estimated costs and salvage value.
11.3 Replacement analysis • Perform a replacementretention study between
a defender and the best challenger.
11.4 Additional considerations • Understand the approach to special situations in
a replacement study.
11.5 Study period analysis • Perform a replacementretention study over a
specified number of years.
11.6 Replacement value • Calculate the minimum market (trade-in) value
required to make the challenger economically
attractive.

O
ne of the most common and important issues in industrial practice is that of replacement
or retention of an asset, process, or system that is currently installed.
This differs from previous situations where all the alternatives were new. The fundamental
question answered by a replacement study (also called a replacementretention
study) about a currently installed system is, Should it be replaced now or later ? When an
asset is currently in use and its function is needed in the future, it will be replaced at some
time. In reality, a replacement study answers the question of when, not if, to replace.
A replacement study is usually designed to first make the economic decision to retain or
replace now . If the decision is to replace, the study is complete. If the decision is to retain,
the cost estimates and decision can be revisited each year to ensure that the decision to
retain is still economically correct. This chapter explains how to perform the initial-year and
follow-on year replacement studies.
A replacement study is an application of the AW method of comparing unequal-life alternatives,
first introduced in Chapter 6. In a replacement study with no specified study period,
the AW values are determined by a technique called the economic service life (ESL) analysis.
If a study period is specified, the replacement study procedure is different from that used
when no study period is set.
If asset depreciation and taxes are to be considered in an after-tax replacement analysis,
Chapters 16 and 17 should be covered before or in conjunction with this chapter. After-tax
replacement analysis is included in Chapter 17.
Keep or Replace the Kiln Case: B&T
Enterprises manufactures and sells highmelting-temperature
ceramics and highperformance
metals to other corporations.
The products are sold to a wide range of
industries from the nuclear and solar power
industry to sports equipment manufacturers
of specialty golf and tennis gear, where kiln
temperatures up to approximately 1700°C
are needed. For years, B&T has owned and
been very satisfied with Harper International
pusher-plate tunnel kilns. Two are in use
currently at plant locations on each coast
of the country; one kiln is 10 years old, and
the second was purchased only 2 years ago
and serves, primarily, the ceramics industry
needs on the west coast. This newer kiln
can reach temperatures of 2800°C.
During the last two or three quarterly
maintenance visits, the Harper team
leader and the head of B&T quality have
discussed the ceramic and metal industry
needs for higher temperatures. In some
cases the temperatures are as high as
3000°C for emerging nitride, boride, and
carbide transition metals that form very
high-melting-temperature oxides. These
may find use in hypersonic vehicles,
engines, plasma arc electrodes, cutting
tools, and high-temperature shielding.
A looming question on the mind of the
senior management and financial officers
of B&T revolves around the need to seriously
consider a new graphite hearth kiln,
which can meet higher temperature and
other needs of the current and projected
PE
customer base. This unit will have lower
operating costs and significantly greater
furnace efficiency in heat time, transit,
and other crucial parameters. Since virtually
all of this business is on the west coast,
the graphite hearth kiln would replace the
newer of the two kilns currently in use.
For identification, let
PT identify the currently installed
pusher-plate tunnel kiln (defender)
GH identify the proposed new graphite
hearth kiln (challenger)
Relevant estimates follow in $ millions
for monetary units.
First cost,
$ M
AOC, $ M
per year
PT
$25; 2 years
ago
year 1: $5.2;
year 2: $6.4
GH
$38; with no
trade-in
starts at $3.4,
increases
10%year
Life, years 6 (remaining) 12 (estimated)
Heating
element,
$ M
—
$2.0 every
6 years
This case is used in the following topics
of this chapter:
Economic service life (Section 11.2)
Replacement study (Section 11.3)
Replacement study with study period
(Section 11.5)
Replacement value (Section 11.6)
Problems 11.18 and 11.41

294 Chapter 11 Replacement and Retention Decisions
11.1 Basics of a Replacement Study
The need for a replacement study can develop from several sources:
Reduced performance. Because of physical deterioration, the ability to perform at an
expected level of reliability (being available and performing correctly when needed) or productivity
(performing at a given level of quality and quantity) is not present. This usually
results in increased costs of operation, higher scrap and rework costs, lost sales, reduced quality,
diminished safety, and larger maintenance expenses.
Altered requirements. New requirements of accuracy, speed, or other specifications cannot
be met by the existing equipment or system. Often the choice is between complete replacement
or enhancement through retrofitting or augmentation.
Obsolescence. International competition and rapidly changing technology make currentlyused
systems and assets perform acceptably but less productively than equipment coming
available. The ever-decreasing development cycle time to bring new products to market is
often the reason for premature replacement studies, that is, studies performed before the
estimated useful or economic life is reached.
Replacement studies use some terminology that is closely related to terms in previous
chapters.
Salvagemarket value
Economic service life
Defender and challenger are the names for two mutually exclusive alternatives. The defender
is the currently installed asset, and the challenger is the potential replacement. A
replacement study compares these two alternatives. The challenger is the “best” challenger
because it has been selected as the best one to possibly replace the defender. (This is the
same terminology used earlier for incremental ROR and BC analysis, but both alternatives
were new).
Market value is the current value of the installed asset if it were sold or traded on the
open market. Also called trade-in value , this estimate is obtained from professional appraisers,
resellers, or liquidators familiar with the industry. As in previous chapters, salvage
value is the estimated value at the end of the expected life. In replacement analysis,
the salvage value at the end of one year is used as the market value at the beginning of the
next year.
AW values are used as the primary economic measure of comparison between the defender
and challenger. The term equivalent uniform annual cost ( EUAC) may be used in lieu of
AW, because often only costs are included in the evaluation; revenues generated by the
defender or challenger are assumed to be equal. (Since EUAC calculations are exactly the
same as for AW, we use the term AW.) Therefore, all values will be negative when only
costs are involved. Salvage or market value is an exception; it is a cash inflow and carries
a plus sign.
Economic service life (ESL) for an alternative is the number of years at which the lowest AW
of cost occurs . The equivalency calculations to determine ESL establish the life n for the best
challenger and the lowest cost life for the defender in a replacement study. The next section
explains how to find the ESL.
Defender first cost is the initial investment amount P used for the defender. The current
market value ( MV ) is the correct estimate to use for P for the defender in a replacement
study. The estimated salvage value at the end of one year becomes the market value at the
beginning of the next year, provided the estimates remain correct as the years pass. It is
incorrect to use the following as MV for the defender first cost: trade-in value that does
not represent a fair market value , or the depreciated book value taken from accounting
records. If the defender must be upgraded or augmented to make it equivalent to the challenger
(in speed, capacity, etc.), this cost is added to the MV to obtain the estimated
defender first cost.
Challenger first cost is the amount of capital that must be recovered (amortized) when
replacing a defender with a challenger. This amount is almost always equal to P , the first cost
of the challenger.

11.1 Basics of a Replacement Study 295
If an unrealistically high trade-in value is offered for the defender compared to its fair
market value, the net cash flow required for the challenger is reduced, and this fact should be
considered in the analysis. The correct amount to recover and use in the economic analysis for
the challenger is its first cost minus the difference between the trade-in value (TIV) and market
value (MV) of the defender. In equation form, this is P (TIV MV). This amount
represents the actual cost to the company because it includes both the opportunity cost (i.e.,
market value of the defender) and the out-of-pocket cost (i.e., first cost trade-in) to acquire
the challenger. Of course, when the trade-in and market values are the same, the challenger P
value is used in all computations.
The challenger first cost is the estimated initial investment necessary to acquire and install it.
Sometimes, an analyst or manager will attempt to increase this first cost by an amount equal to the
unrecovered capital remaining in the defender, as shown on the accounting records for the asset.
This incorrect treatment of capital recovery is observed most often when the defender is working
well and in the early stages of its life, but technological obsolescence, or some other reason, has
forced consideration of a replacement. This leads us to identify two additional characteristics of
replacement analysis, in fact, of any economic analysis: sunk costs and nonowner’s viewpoint .
A sunk cost is a prior expenditure or loss of capital (money) that cannot be recovered by a
decision about the future. The replacement alternative for an asset, system, or process that has
incurred a nonrecoverable cost should not include this cost in any direct fashion; sunk costs
should be handled in a realistic way using tax laws and write-off allowances.
Sunk cost
A sunk cost should never be added to the challenger’s first cost, because it will make the challenger
appear to be more costly than it actually is . For example, assume an asset costing $100,000
two years ago has a depreciated value of $80,000 on the corporate books. It must be replaced
prematurely due to rapidly advancing technology. If the replacement alternative (challenger) has
a first cost of $150,000, the $80,000 from the current asset is a sunk cost were the challenger
purchased. For the purposes of an economic analysis, it is incorrect to increase the challenger’s
first cost to $230,000 or any number between this and $150,000.
The second characteristic is the perspective taken when conducting a replacement study. You,
the analyst, are a consultant from outside the company.
The nonowner’s viewpoint , also called the outsider’s viewpoint or consultant’s viewpoint, provides
the greatest objectivity in a replacement study. This viewpoint performs the analysis without
bias; it means the analyst owns neither the defender nor the challenger. Additionally, it assumes
the services provided by the defender can be purchased now by making an “initial
investment” equal to the market value of the defender.
Besides being unbiased, this perspective is correct because the defender’s market value is a forgone
opportunity of cash inflow were the replacement not selected, and the defender chosen.
As mentioned in the introduction, a replacement study is an application of the annual worth
method. As such, the fundamental assumptions for a replacement study parallel those of an AW
analysis. If the planning horizon is unlimited , that is, a study period is not specified, the assumptions
are as follows:
1. The services provided are needed for the indefinite future.
2. The challenger is the best challenger available now and in the future to replace the defender.
When this challenger replaces the defender (now or later), it will be repeated for succeeding
life cycles.
3. Cost estimates for every life cycle of the defender and challenger will be the same as in their
first cycle.
As expected, none of these assumptions is precisely correct. We discussed this previously for the
AW method (and the PW method). When the intent of one or more of the assumptions becomes
incorrect, the estimates for the alternatives must be updated and a new replacement study conducted.
The replacement procedure discussed in Section 11.3 explains how to do this. When the
planning horizon is limited to a specified study period, the assumptions above do not hold. The
procedure of Section 11.5 discusses replacement analysis over a fixed study period.

296 Chapter 11 Replacement and Retention Decisions
EXAMPLE 11.1
Only 2 years ago, Techtron purchased for $275,000 a fully loaded SCADA (supervisory control
and data acquisition) system including hardware and software for a processing plant operating
on the Houston ship channel. When it was purchased, a life of 5 years and salvage of 20%
of first cost were estimated. Actual M&O costs have been $25,000 per year, and the book value
is $187,000. There has been a series of insidious malware infections targeting Techtron’s command
and control software, plus next-generation hardware marketed only recently could
greatly reduce the competitiveness of the company in several of its product lines. Given these
factors, the system is likely worth nothing if kept in use for the final 3 years of its anticipated
useful life.
Model K2-A1, a new replacement turnkey system, can be purchased for $300,000 net cash,
that is, $400,000 first cost and a $100,000 trade-in for the current system. A 5-year life, salvage
value of 15% of stated first cost or $60,000, and an M&O cost of $15,000 per year are good
estimates for the new system. The current system was appraised this morning, and a market
value of $100,000 was confirmed for today; however, with the current virus discovery, the appraiser
anticipates that the market value will fall rapidly to the $80,000 range once the virus
problem and new model are publicized.
Using the above values as the best possible today, state the correct defender and challenger
estimates for P, M&O, S, and n in a replacement study to be performed today.
Solution
Defender: Use the current market value of $100,000 as the first cost for the defender. All
others—original cost of $275,000, book value of $187,000, and trade-in value of
$100,000—are irrelevant to a replacement study conducted today. The estimates are as
follows:
First cost P $100,000
M&O cost A $25,000 per year
Expected life n 3 years
Salvage value S 0
Challenger: The $400,000 stated first cost is the correct one to use for P, because the
trade-in and market values are equal.
First cost P $400,000
M&O cost A $15,000 per year
Expected life n 5 years
Salvage value S $60,000
Comment
If the replacement study is conducted next week when estimates will have changed, the defender’s
first cost will be $80,000, the new market value according to the appraiser. The challenger’s
first cost will be $380,000, that is, P (TIV MV) 400,000 (100,000 – 80,000).
11.2 Economic Service Life
Until now the estimated life n of an alternative or asset has been stated. In reality, the best life
estimate to use in the economic analysis is not known initially. When a replacement study or an
analysis between new alternatives is performed, the best value for n should be determined using
current cost estimates. The best life estimate is called the economic service life .
Economic service life
The economic service life (ESL) is the number of years n at which the equivalent uniform
annual worth (AW) of costs is the minimum, considering the most current cost estimates over
all possible years that the asset may provide a needed service.

11.2 Economic Service Life 297
Larger
costs
Figure 11–1
Annual worth curves of
cost elements that determine
the economic
service life.
AW of costs, $/year
Total AW of costs
AW of AOC
Capital recovery
0 Years
Economic
service life
The ESL is also referred to as the economic life or minimum cost life . Once determined, the ESL
should be the estimated life for the asset used in an engineering economy study, if only economics
are considered. When n years have passed, the ESL indicates that the asset should be replaced
to minimize overall costs. To perform a replacement study correctly, it is important that the ESL
of the challenger and the ESL of the defender be determined, since their n values are usually not
preestablished.
The ESL is determined by calculating the total AW of costs if the asset is in service 1 year,
2 years, 3 years, and so on, up to the last year the asset is considered useful. Total AW of costs is
the sum of capital recovery (CR), which is the AW of the initial investment and any salvage
value, and the AW of the estimated annual operating cost (AOC), that is,
Total AW capital recovery AW of annual operating costs
CR AW of AOC [11.1]
The ESL is the n value for the smallest total AW of costs . (Remember: These AW values are cost
estimates, so the AW values are negative numbers. Therefore, $–200 is a lower cost than $500.)
Figure 11–1 shows the characteristic shape of a total AW of cost curve. The CR component of total
AW decreases, while the AOC component increases, thus forming the concave shape. The two AW
components are calculated as follows.
Decreasing cost of capital recovery. The capital recovery is the AW of investment; it decreases
with each year of ownership. Capital recovery is calculated by Equation [6.3], which
is repeated here. The salvage value S , which usually decreases with time, is the estimated
market value (MV) in that year.
Capital recovery P ( AP , i,n ) S ( A F , i,n ) [11.2]
Increasing cost of AW of AOC. Since the AOC (or M&O) estimates usually increase over the
years, the AW of AOC increases. To calculate the AW of the AOC series for 1, 2, 3, . . . years,
determine the present worth of each AOC value with the PF factor, then redistribute this
P value over the years of ownership, using the AP factor.
The complete equation for total AW of costs over k years (k 1, 2, 3, . . . ) is
jk
Total AW k P(AP,i,k) S k (AF,i,k)
AOC j (PF,i,j) (AP,i,k) [11.3]
j1
Capital recovery
where
P initial investment or current market value
S k salvage value or market value after k years
AOC j annual operating cost for year j ( j 1 to k )

298 Chapter 11 Replacement and Retention Decisions
The current MV is used for P when the asset is the defender, and the estimated future MV values
are substituted for the S values in years 1, 2, 3, . . . . Plotting the AW k series as in Figure 11–1
clearly indicates where the ESL is located and the trend of the AW k curve on each side of the ESL.
To determine ESL by spreadsheet, the PMT function (with embedded NPV functions as
needed) is used repeatedly for each year to calculate capital recovery and the AW of AOC. Their
sum is the total AW for k years of ownership. The PMT function formats for the capital recovery
and AOC components for each year k ( k 1, 2, 3, . . .) are as follows:
Capital recovery for the challenger:
Capital recovery for the defender:
AW of AOC:
PMT( i %,years, P, −MV_in_year_k)
PMT( i %,years,current_MV,MV_in_
year_k) [11.4]
PMT( i %,years,NPV( i %,year_1_AOC:
year_k_AOC)0)
EXAMPLE 11.2
When the spreadsheet is developed, it is recommended that the PMT functions in year 1 be developed
using cell-reference format; then drag down the function through each column. A final
column summing the two PMT results displays total AW. Augmenting the table with an Excel xy
scatter chart graphically displays the cost curves in the general form of Figure 11–1 , and the ESL
is easily identified. Example 11.2 illustrates ESL determination by hand and by spreadsheet.
A 3-year-old backup power system is being considered for early replacement. Its current market
value is $20,000. Estimated future market values and annual operating costs for the next
5 years are given in Table 11–1, columns 2 and 3. What is the economic service life of this
defender if the interest rate is 10% per year? Solve by hand and by spreadsheet.
Solution by Hand
Equation [11.3] is used to calculate total AW k for k 1, 2, . . . , 5. Table 11–1, column 4, shows
the capital recovery for the $20,000 current market value ( j 0) plus 10% return. Column 5
gives the equivalent AW of AOC for k years. As an illustration, the computation of total AW
for k 3 from Equation [11.3] is
Total AW 3 P(AP,i,3) MV 3 (AF,i,3) [PW of AOC 1 ,AOC 2 , and AOC 3 ](AP,i,3)
20,000(AP,10%,3) 6000(AF,10%,3) [5000(PF,10%,1)
6500(PF,10%,2) 8000(PF,10%,3)](AP,10%,3)
6230 6405 $12,635
A similar computation is performed for each year 1 through 5. The lowest equivalent cost
( numerically largest AW value) occurs at k 3. Therefore, the defender ESL is n 3 years,
and the AW value is $12,635. In the replacement study, this AW will be compared with the
best challenger AW determined by a similar ESL analysis.
Solution by Spreadsheet
See Figure 11–2 for the spreadsheet screen shot and chart that shows the ESL is n 3 years
and AW $12,634. (This format is a template for any ESL analysis; simply change the
TABLE 11–1
Computation of Economic Service Life
Year j
(1)
MV j , $
(2)
AOC j , $
(3)
Capital
Recovery, $
(4)
AW of
AOC, $
(5)
Total
AW k , $
(6) (4) (5)
1 10,000 5,000 12,000 5,000 17,000
2 8,000 6,500 7,714 5,714 13,428
3 6,000 8,000 6,230 6,405 12,635
4 2,000 9,500 5,878 7,072 12,950
5 0 12,500 5,276 7,961 13,237

11.2 Economic Service Life 299
Current market value
PMT($B$1,$A9,$B$2,$B9)
PMT($B$1,$A9,NPV($B$1,$C$5:$C9)0)
ESL of defender
Total AW is minimum
at n 3 years
Total AW
curve
Capital
recovery
curve
Figure 11–2
Determination of ESL and plot of curves, Example 11.2
estimates and add rows for more years.) Contents of columns D and E are described below. The
PMT functions apply the formats as described in Equation [11.4]. Cell tags show detailed cellreference
format for year 5. The $ symbols are included for absolute cell referencing, needed
when the entry is dragged down through the column.
Column D: Capital recovery is the AW of the $20,000 investment in year 0 for each year 1
through 5 with the estimated MV in that year. For example, in actual numbers, the
cell-reference PMT function in year 5 shown on the spreadsheet reads PMT (10%,5,20000,
0), resulting in $5276. This series is plotted in Figure 11–2.
Column E: The NPV function embedded in the PMT function obtains the present worth in
year 0 of all AOC estimates through year k. Then PMT calculates the AW of AOC over
k years. For example, in year 5, the PMT in numbers is PMT(10%,5,NPV
(10%,C5:C9)0). The 0 is the AOC in year 0; it is optional. The graph plots the AW of
AOC curve, which constantly increases in cost because the AOC estimates increase
each year.
Comment
The capital recovery curve in Figure 11–2 (middle curve) is not the expected shape (see year 4)
because the estimated market value changes each year. If the same MV were estimated for each
year, the curve would appear like Figure 11–1. When several total AW values are approximately
equal, the curve will be flat over several periods. This indicates that the ESL is relatively
insensitive to costs.

300 Chapter 11 Replacement and Retention Decisions
It is reasonable to ask about the difference between the ESL analysis above and the AW
analyses performed in previous chapters. Previously we had a specifi c life estimated to be n years
with associated other estimates: first cost in year 0, possibly a salvage value in year n , and an
AOC that remained constant or varied each year. For all previous analyses, the calculation of AW
using these estimates determined the AW over n years. This is the economic service life when n
is fixed. Also, in all previous cases, there were no year-by-year market value estimates. Therefore,
we can conclude the following:
When the expected life n is known and specified for the challenger or defender, no ESL computations
are necessary. Determine the AW over n years, using the first cost or current market
value, estimated salvage value after n years, and AOC estimates. This AW value is the correct one
to use in the replacement study.
However, when n is not fixed, the following is useful. First the market/salvage series is needed.
It is not difficult to estimate this series for a new or current asset. For example, an asset with a
first cost of P can lose market value of, say, 20% per year, so the market value series for years
0, 1, 2, . . . is P , 0.8 P , 0.64 P , . . . , respectively. If it is reasonable to predict the MV series on a
year-by-year basis, it can be combined with the AOC estimates to produce what is called the
marginal costs for the asset.
Marginal costs (MC) are year-by-year estimates of the costs to own and operate an asset for that
year. Three components are added to determine the marginal cost:
• Cost of ownership (loss in market value is the best estimate of this cost)
• Forgone interest on the market value at the beginning of the year
• AOC for each year
Once the marginal costs are estimated for each year, their equivalent AW value is calculated. The
sum of the AW values of the fi rst two of these components is the capital recovery amount . Now, it
should be clear that the total AW of all three marginal cost components over k years is the same
value as the total annual worth for k years calculated in Equation [11.3]. That is, the following
relation is correct.
AW of marginal costs total AW of costs [11.5]
Therefore, there is no need to perform a separate, detailed marginal cost analysis when yearly
market values are estimated. The ESL analysis presented in Example 11.2 is sufficient in that
it results in the same numerical values. This is demonstrated in Example 11.3 using the progressive
example.
EXAMPLE 11.3 Keep or Replace the Kiln Case
PE
In our progressive example, B&T Enterprises is considering the replacement of a 2-year-old
kiln with a new one to meet emerging market needs. When the current tunnel kiln was purchased
2 years ago for $25 million, an ESL study indicated that the minimum cost life was between
3 and 5 years of the expected 8-year life. The analysis was not very conclusive because
the total AW cost curve was flat for most years between 2 and 6, indicating insensitivity of the
ESL to changing costs. Now, the same type of question arises for the proposed graphite hearth
model that costs $38 million new: What are the ESL and the estimated total AW of costs? The
Manager of Critical Equipment at B&T estimates that the market value after only 1 year will
drop to $25 million and then retain 75% of the previous year’s value over the 12-year expected
life. Use this market value series and i 15% per year to illustrate that an ESL analysis and
marginal cost analysis result in exactly the same total AW of cost series.
Solution
Figure 11–3 is a spreadsheet screen shot of the two analyses in $ million units. The market
value series is detailed in column B starting at $25 (million) and decreasing by 25% per year.
A brief description of each analysis follows.

11.2 Economic Service Life 301
ESL analysis
Two AW series
are identical
Marginal cost analysis
Figure 11–3
Comparison of annual worth series resulting from ESL analysis and marginal cost analysis, Example 11.3.
ESL analysis: Equation [11.4] is applied repeatedly for k 1, 2, . . . , 12 years (columns C,
D, and E) in the top of Figure 11–3. Row 16 details the spreadsheet functions for year 12.
The result in column F is the total AW series that is of interest now.
Marginal cost (MC): The functions in the bottom of Figure 11–3 (columns C, D, and E)
develop the three components added to obtain the MC series. Row 33 details the functions
for year 12. The resulting AW of marginal costs (column G) is the series to compare with
the corresponding ESL series above (column F).
The two AW series are identical, thus demonstrating that Equation [11.5] is correct. Therefore,
either an ESL or a marginal cost analysis will provide the same information for a replacement
study. In this case, the results show that the new kiln will have a minimum AW of costs
of $–12.32 million at its full 12-year life.
We can draw two important conclusions about the n and AW values to be used in a replacement
study. These conclusions are based on the extent to which detailed annual estimates are
made for the market value.
1. Year-by-year market value estimates are made. Use them to perform an ESL analysis,
and determine the n value with the lowest total AW of costs. These are the best n and AW
values for the replacement study.
2. Yearly market value estimates are not available. The only estimate available is market
value (salvage value) in year n . Use it to calculate the AW over n years. These are
the n and AW values to use; however, they may not be the “best” values in that they may
not represent the best equivalent total AW of cost value found if an ESL analysis were
performed.

302 Chapter 11 Replacement and Retention Decisions
Upon completion of the ESL analysis (item 1 above), the replacement study procedure in Section
11.3 is applied using the values
Challenger alternative (C):
Defender alternative (D):
AW C for n C years
AW D for n D years
11.3 Performing a Replacement Study
Replacement studies are performed in one of two ways: without a study period specified or with
one defined. Figure 11–4 gives an overview of the approach taken for each situation. The procedure
discussed in this section applies when no study period (planning horizon) is specified. If a
specific number of years is identified for the replacement study, for example, over the next
5 years, with no continuation considered after this time period in the economic analysis, the procedure
in Section 11.5 is applied.
A replacement study determines when a challenger replaces the in-place defender. The complete
study is finished if the challenger (C) is selected to replace the defender (D) now. However,
if the defender is retained now, the study may extend over a number of years equal to the life of
the defender n D , after which a challenger replaces the defender. Use the annual worth and life
values for C and D determined in the ESL analysis in the following procedure. Assume the services
provided by the defender could be obtained at the AW D amount.
The replacement study procedure is:
New replacement study:
1. On the basis of the better AW C or AW D value, select the challenger C or defender D. When the
challenger is selected, replace the defender now, and expect to keep the challenger for n C
years. This replacement study is complete. If the defender is selected, plan to retain it for up
to n D more years. (This is the leftmost branch of Figure 11–4 .) Next year, perform the
following steps.
One-year-later analysis:
2. Determine if all estimates are still current for both alternatives, especially first cost, market
value, and AOC. If not, proceed to step 3. If yes and this is year n D , replace the defender. If
this is not year n D , retain the defender for another year and repeat this same step. This step
may be repeated several times.
3. Whenever the estimates have changed, update them and determine new AW C and AW D values.
Initiate a new replacement study (step 1).
Figure 11–4
Overview of replacement
study approaches.
Replacement study
No study period
specified
Study period
specified
Perform ESL
analysis
Develop succession
options for D and C
using AW of
respective cash flows
AW D
AW C
PW or AW for
each option
Select better AW
Select best option

11.3 Performing a Replacement Study 303
If the defender is selected initially (step 1), estimates may need updating after 1 year of retention
(step 2). Possibly there is a new best challenger to compare with D. Either significant
changes in defender estimates or availability of a new challenger indicates that a new replacement
study is to be performed. In actuality, a replacement study can be performed each year or
more frequantly to determine the advisability of replacing or retaining any defender, provided
a competitive challenger is available.
Example 11.4 illustrates the application of ESL analysis for a challenger and defender, followed
by the use of the replacement study procedure. The planning horizon is unspecified in this
example.
EXAMPLE 11.4
Two years ago, Toshiba Electronics made a $15 million investment in new assembly line
machinery. It purchased approximately 200 units at $70,000 each and placed them in plants
in 10 different countries. The equipment sorts, tests, and performs insertion-order kitting on
electronic components in preparation for special-purpose circuit boards. This year, new international
industry standards will require a $16,000 retrofit on each unit, in addition to the expected
operating cost. Due to the new standards, coupled with rapidly changing technology, a
new system is challenging the retention of these 2-year-old machines. The chief engineer at
Toshiba USA realizes that the economics must be considered, so he has asked that a replacement
study be performed this year and each year in the future, if need be. The i is 10% and the
estimates are below.
Challenger: First cost: $50,000
Future market values: decreasing by 20% per year
Estimated retention period: no more than 10 years
AOC estimates: $5000 in year 1 with increases of $2000 per year
thereafter
Defender: Current international market value: $15,000
Future market values: decreasing by 20% per year
Estimated retention period: no more than 3 more years
AOC estimates: $4000 next year, increasing by $4000 per year
thereafter, plus the $16,000 retrofit next year
(a) Determine the AW values and economic service lives necessary to perform the replacement
study.
(b) Perform the replacement study now.
(c) After 1 year, it is time to perform the follow-up analysis. The challenger is making large
inroads to the market for electronic components assembly equipment, especially with the
new international standards features built in. The expected market value for the defender is
still $12,000 this year, but it is expected to drop to virtually nothing in the future—$2000
next year on the worldwide market and zero after that. Also, this prematurely outdated
equipment is more costly to keep serviced, so the estimated AOC next year has been increased
from $8000 to $12,000 and to $16,000 two years out. Perform the follow-up replacement
study analysis.
Solution
(a) The results of the ESL analysis, shown in Figure 11–5, include all the MV and AOC estimates
in columns B and C. For the challenger, note that P $50,000 is also the MV in
year 0. The total AW of costs is for each year, should the challenger be placed into service
for that number of years. As an example, the year k 4 amount of $19,123 is determined
using Equation [11.3], where the AG factor accommodates the arithmetic gradient
series in the AOC.
Total AW 4 50,000(AP,10%,4) 20,480(AF,10%,4)
[5000 2000(AG,10%,4)]
$19,123

304 Chapter 11 Replacement and Retention Decisions
ESL Analysis of Challenger
ESL results
ESL Analysis of Defender
Figure 11–5
ESL analysis of challenger and defender, Example 11.4.
For spreadsheet-based ESL analysis, this same result is achieved in cell F8 using Equation
[11.4]. The functions are
Total AW 4 PMT(10%,4,50000,20480) PMT(10%,4,NPV(10%,C5:C8)0)
11,361 7,762
19,123
The defender costs are analyzed in the same way up to the maximum retention period of
3 years.
The lowest AW cost (numerically largest) values for the replacement study are as follows:
Challenger: AW C $19,123 for n C 4 years
Defender: AW D $17,307 for n D 3 years
The challenger total AW of cost curve (Figure 11–5) is classically shaped and relatively flat
between years 3 and 6; there is virtually no difference in the total AW for years 4 and 5. For
the defender, note that the estimated AOC values change substantially over 3 years, and
they do not constantly increase or decrease.
(b) To perform the replacement study now, apply only the first step of the procedure. Select the
defender because it has the better AW of costs ($17,307), and expect to retain it for
3 more years. Prepare to perform the one-year-later analysis 1 year from now.
(c) One year later, the situation has changed significantly for the equipment Toshiba retained
last year. Apply the steps for the one-year-later analysis:
2. After 1 year of defender retention, the challenger estimates are still reasonable, but the
defender market value and AOC estimates are substantially different. Go to step 3 to
perform a new ESL analysis for the defender.
3. The defender estimates in Figure 11–5 are updated below for the ESL analysis. New
AW values are calculated using Equation [11.3]. There is now a maximum of 2 more
years of retention, 1 year less than the 3 years determined last year.
Year k
Market
Value, $ AOC, $
Total AW
If Retained k More Years, $
0 12,000 — —
1 2,000 12,000 23,200
2 0 16,000 20,819

11.3 Performing a Replacement Study 305
The AW and n values for the new replacement study are as follows:
Challenger:
Defender:
unchanged at AW C $19,123 for n C 4 years
new AW D $20,819 for n D 2 more years
Now select the challenger based on its favorable AW value. Therefore, replace the defender
now, not 2 years from now. Expect to keep the challenger for 4 years, or until a better challenger
appears on the scene.
EXAMPLE 11.5 Keep or Replace the Kiln Case
We continue with the progressive example of possibly replacing a kiln at B&T Enterprises. A
marketing study revealed that the improving business activity on the west coast implies that
the revenue profile between the installed kiln (PT) and the proposed new one (GH) would be
the same, with the new kiln possibly bringing in new revenue within the next couple of years.
The president of B&T decided it was time to do a replacement study. Assume you are the lead
engineer and that you previously completed the ESL analysis on the challenger (Example
11.3). It indicates that for the GH system the ESL is its expected useful life.
Challenger: ESL n GH 12 years with total equivalent annual cost AW GH $12.32 million
The president asked you to complete the replacement study, stipulating that, due to the rapidly
rising annual operating costs (AOC), the defender would be retained a maximum of 6 years.
You are expected to make the necessary estimates for the defender (PT) and perform the study
at a 15% per year return.
PE
Solution
After some data collection, you have good evidence that the market value for the PT system
will stay high, but that the increasing AOC is expected to continue rising about $1.2 million per
year. The best estimates for the next 6 years in $ million units are these:
Year 1 2 3 4 5 6
Market Value, $ M 22.0 22.0 22.0 20.0 18.0 18.0
AOC, $ M per year 5.2 6.4 7.6 8.8 10.0 11.2
You developed a spreadsheet and performed the analysis in Figure 11–6. As an illustration,
total AW computation for 3 years of retention, in $ million units, is
Total AW 3 22.0(AP,15%,3) 22.0(AF,15%,3) [5.2(PF,15%,1) 6.4(PF,15%,2)
7.6(PF,15%,3)](AP,15%,3)
–9.63 6.34 [14.36] (0.43798)
$9.59 per year
Though the system could be retained up to 6 years, the ESL is much shorter at 1 year.
Defender: ESL n PT 1 year with total equivalent annual cost AW PT $8.50 million
Figure 11–6
ESL analysis of defender kiln PT for progressive example, Example 11.5.

306 Chapter 11 Replacement and Retention Decisions
To make the replacementretention decision, apply step 1 of the procedure. Since AW PT
$8.50 million per year is considerably less than AW $12.32 million, you should recommend
keeping the current kiln only 1 more year and doing another study during the year
to determine if the current estimates are still reliable.
Comment
An observation of the trends of the two final AW series in this problem is important. A comparison
of Figure 11–3 (top), column F, and Figure 11–6, column F, shows us that the largest
total AW of the current system ($–11.47 M for 6 years) is still below the smallest total AW of
the proposed system ($12.32 M for 12 years). This indicates that the graphite hearth system
(the challenger) will not be chosen on an economic basis, if the decision to consider the installed
kiln as a defender in the future were made. It would take some significant estimate
changes to justify the challenger.
11.4 Additional Considerations in a
Replacement Study
There are several additional aspects of a replacement study that may be introduced. Three of
these are identified and discussed in turn.
• Future-year replacement decisions at the time of the initial replacement study
• Opportunity cost versus cash flow approaches to alternative comparison
• Anticipation of improved future challengers
In most cases when management initiates a replacement study, the question is best framed as,
“Replace now, 1 year from now, 2 years from now, etc.?” The procedure above does answer this
question provided the estimates for C and D do not change as each year passes. In other words,
at the time it is performed, step 1 of the procedure does answer the replacement question for
multiple years. It is only when estimates change over time that the decision to retain the defender
may be prematurely reversed (prior to n D years) in favor of the then-best challenger.
The first costs ( P values) for the challenger and defender have been correctly taken as the
initial investment for the challenger C and current market value for the defender D. This is called
the opportunity cost approach because it recognizes that a cash inflow of funds equal to the
market value is forgone if the defender is selected. This approach, also called the conventional
approach, is correct for every replacement study. A second approach, called the cash flow approach,
recognizes that when C is selected, the market value cash inflow for the defender is received
and, in effect, immediately reduces the capital needed to invest in the challenger. Use of
the cash fl ow approach is strongly discouraged for at least two reasons: possible violation of the
equal-service requirement and incorrect capital recovery value for C. As we are aware, all economic
evaluations must compare alternatives with equal service. Therefore, the cash flow approach
can work only when challenger and defender lives are exactly equal. This is commonly
not the case; in fact, the ESL analysis and the replacement study procedure are designed to compare
two mutually exclusive, unequal-life alternatives via the annual worth method. If this equalservice
comparison reason is not enough to avoid the cash flow approach, consider what happens
to the challenger’s capital recovery amount when its first cost is decreased by the market value of
the defender. The capital recovery (CR) terms in Equation [11.3] will decrease, resulting in a
falsely low value of CR for the challenger, were it selected. From the vantage point of the economic
study itself, the decision for C or D will not change; but when C is selected and implemented,
this CR value is not reliable. The conclusion is simple:
Use the initial investment of C and the market value of D as the first costs in the ESL analysis and
in the replacement study.
A basic premise of a replacement study is that some challenger will replace the defender at a future
time, provided the service continues to be needed and a worthy challenger is available. The expectation
of ever-improving challengers can offer strong encouragement to retain the defender until
some situational elements—technology, costs, market fluctuations, contract negotiations, etc.—

11.5 Replacement Study over a Specified Study Period 307
stabilize. This was the case in the previous two examples. A large expenditure on equipment when
the standards changed soon after purchase forced an early replacement consideration and a large loss
of invested capital. The replacement study is no substitute for forecasting challenger availability. It is
important to understand trends, new advances, and competitive pressures that can complement the
economic outcome of a good replacement study . It is often better to compare a challenger with an
augmented defender in the replacement study. Adding needed features to a currently installed defender
may prolong its useful life and productivity until challenger choices are more appealing.
It is possible that a significant tax impact may occur when a defender is traded early in its
expected life. If taxes should be considered, proceed now, or after the next section, to Chapter 17
and the after-tax replacement analysis in Section 17.7.
11.5 Replacement Study over a
Specified Study Period
When the time period for the replacement study is limited to a specified study period or planning
horizon, for example, 6 years, the ESL analysis is not performed.
The AW values for the challenger and for the remaining life of the defender are not based on the
economic service life; the AW is calculated over the study period only. What happens to the alternatives
after the study period is not considered in the replacement analysis.
This means that the defender or challenger is not needed beyond the study period. In fact,
a study period of fixed duration does not comply with the three assumptions stated in
Section 11.1—service needed for indefinite future, best challenger available now, and estimates
will be identical for future life cycles.
When performing a replacement study over a fixed study period, it is crucial that the estimates
used to determine the AW values be accurate and used in the study. This is especially important for
the defender. Failure to do the following violates the requirement of equal-service comparison.
When the defender’s remaining life is shorter than the study period , the cost of providing
the defender’s services from the end of its expected remaining life to the end of the study
period must be estimated as accurately as possible and included in the replacement study.
Study period
The right branch of Figure 11–4 presents an overview of the replacement study procedure for a
stated study period.
1. Succession options and AW values . Develop all the viable ways to use the defender and challenger
during the study period. There may be only one option or many options; the longer the
study period, the more complex this analysis becomes. The AW values for the challenger and
defender cash flows are used to build the equivalent cash flow values for each option.
2. Selection of the best option . The PW or AW for each option is calculated over the study
period. Select the option with the lowest cost, or highest income if revenues are estimated.
(As before, the best option will have the numerically largest PW or AW value.)
The following examples use this procedure and illustrate the importance of making cost estimates
for the defender alternative when its remaining life is less than the study period.
EXAMPLE 11.6
Claudia works with Lockheed-Martin (LMCO) in the aircraft maintenance division. She is
preparing for what she and her boss, the division chief, hope to be a new 10-year defense contract
with the U.S. Air Force on C-5A cargo aircraft. A key piece of equipment for maintenance
operations is an avionics circuit diagnostics system. The current system was purchased 7 years
ago on an earlier contract. It has no capital recovery costs remaining, and the following are
reliable estimates: current market value $70,000, remaining life of 3 more years, no salvage
value, and AOC $30,000 per year. The only options for this system are to replace it now or
retain it for the full 3 additional years.
Claudia has found that there is only one good challenger system. Its cost estimates are: first
cost $750,000, life 10 years, S 0, and AOC $50,000 per year.

308 Chapter 11 Replacement and Retention Decisions
Realizing the importance of accurate defender alternative cost estimates, Claudia asked the
division chief what system would be a logical follow-on to the current one 3 years hence, if
LMCO wins the contract. The chief predicted LMCO would purchase the very system she had
identified as the challenger, because it is the best on the market. The company would keep it
for the entire 10 additional years for use on an extension of this contract or some other application
that could recover the remaining 3 years of invested capital. Claudia interpreted the response
to mean that the last 3 years would also be capital recovery years, but on some project
other than this one. Claudia’s estimate of the first cost of this same system 3 years from now is
$900,000. Additionally, the $50,000 per year AOC is the best estimate at this time.
The division chief mentioned any study had to be conducted using the interest rate of 10%,
as mandated by the U.S. Office of Management and Budget (OMB). Perform a replacement
study for the fixed contract period of 10 years.
Solution
The study period is fixed at 10 years, so the intent of the replacement study assumptions is not
present. This means the defender follow-on estimates are very important to the analysis.
Further, any analyses to determine the ESL values are unnecessary since alternative lives are
already set and no projected annual market values are available. The first step of the replacement
study procedure is to define the options. Since the defender will be replaced now or in
3 years, there are only two options:
1. Challenger for all 10 years.
2. Defender for 3 years, followed by the challenger for 7 years.
Cash flows are diagrammed in Figure 11–7. For option 1, the challenger is used for all 10 years.
Equation [11.3] is applied to calculate AW using the following estimates:
Challenger: P $750,000 AOC $50,000
n 10 years S 0
AW C 750,000(AP,10%,10) 50,000 $172,063
0 1 2 3 4 5 6 7 8 9 10
Year
P = $750,000
AOC = $50,000
(a) Challenger (option 1)
End of
study period
0 1 2 3 4 5 6 7 8 9 10 11 12 13
Year
AOC = $30,000
MV = $70,000
Defendercurrent
AOC = $50,000
CR = $146,475
Defender-follow-on
CR for another
project
(b) Defender alternative (option 2)
Figure 11–7
Cash flow diagrams for a 10-year study period replacement study, Example 11.6.

11.5 Replacement Study over a Specified Study Period 309
The second option has more complex cost estimates. The AW for the in-place system is calculated
over the first 3 years. Added to this is the capital recovery for the defender follow-on for the
next 7 years. However in this case, the CR amount is determined over its full 10-year life. (It is not
unusual for the recovery of invested capital to be moved between projects, especially for contract
work.) Refer to the AW components as AW DC (subscript DC for defender current) and AW DF (subscript
DF for defender follow-on). The final cash flows are shown in Figure 11–7b.
Defender current: market value $70,000 AOC $30,000
n 3 years S 0
AW DC [70,000 30,000(PA,10%,3)](AP,10%,10) $23,534
Defender follow-on: P $900,000, n 10 years for capital recovery calculation
only, AOC $–50,000 for years 4 through 10, S 0.
The CR and AW for all 10 years are
CR DF 900,000(AP,10%,10) $146,475 [11.6]
AW DF (146,475 50,000)(FA,10%,7)(AF,10%,10) $116,966
Total AW D for the defender is the sum of the two annual worth values above. This is the AW
for option 2.
AW D AW DC AW DF 23,534 116,966 $140,500
Option 2 has a lower cost ($140,500 versus $172,063). Retain the defender now and expect
to purchase the follow-on system 3 years hence.
Comment
The capital recovery cost for the defender follow-on will be borne by some yet-to-be-identified
project for years 11 through 13. If this assumption were not made, its capital recovery cost would
be calculated over 7 years, not 10, in Equation [11.6], increasing CR to $184,869. This raises the
annual worth to AW D $163,357. The defender alternative (option 2) is still selected.
EXAMPLE 11.7
Three years ago Chicago’s O’Hare Airport purchased a new fire truck. Because of flight increases,
new fire-fighting capacity is needed once again. An additional truck of the same capacity
can be purchased now, or a double-capacity truck can replace the current fire truck. Estimates
are presented below. Compare the options at 12% per year using (a) a 12-year study
period and (b) a 9-year study period.
Presently Owned New Purchase Double Capacity
First cost P, $ 151,000 (3 years ago) 175,000 190,000
AOC, $ 1,500 1,500 2,500
Market value, $ 70,000 — —
Salvage value, $ 10% of P 12% of P 10% of P
Life, years 12 12 12
Solution
Identify option 1 as retention of the presently owned truck and augmentation with a new samecapacity
vehicle. Define option 2 as replacement with the double-capacity truck.
Option 1 Option 2
Presently Owned Augmentation Double Capacity
P, $ 70,000 175,000 190,000
AOC, $ 1,500 1,500 2,500
S, $ 15,100 21,000 19,000
n, years 9 12 12

310 Chapter 11 Replacement and Retention Decisions
(a) For a full-life 12-year study period of option 1,
AW 1 (AW of presently owned) (AW of augmentation)
[70,000(AP,12%,9) 15,100(AF,12%,9) 1500]
[175,000(AP,12%,12) 21,000(AF,12%,12) 1500]
13,616 28,882
$42,498
This computation assumes the equivalent services provided by the current fire truck can be
purchased at $−13,616 per year for years 10 through 12.
AW 2 190,000(AP,12%,12) 19,000(AF,12%,12) 2500
$32,386
Replace now with the double-capacity truck (option 2) at an advantage of $10,112 per year.
(b) The analysis for an abbreviated 9-year study period is identical, except that n 9 in each
factor; that is, 3 fewer years are allowed for the augmentation and double-capacity trucks
to recover the capital investment plus a 12% per year return. The salvage values remain the
same since they are quoted as a percentage of P for all years.
Option 2 is again selected.
AW 1 $46,539
AW 2 $36,873
The previous two examples indicate an important consideration for setting the length of the
study period for a replacement analysis. It involves the capital recovery amount for the challenger,
when the strict definition of a study period is applied.
Study period
Capital recovery
When a study period shorter than the life of the challenger is defined, the challenger’s capital
recovery amount increases in order to recover the initial investment plus a return in this
shortened time period . Highly abbreviated study periods tend to disadvantage the challenger
because no consideration of time beyond the end of the study period is made in calculating the
challenger’s capital recovery amount.
If there are several options for the number of years that the defender may be retained before
replacement with the challenger, the first step of the replacement study—succession options and
AW values—must include all the viable options. For example, if the study period is 5 years and
the defender will remain in service 1 year, or 2 years, or 3 years, cost estimates must be made to
determine AW values for each defender retention period. In this case, there are four options; call
them W, X, Y, and Z.
Option
Defender
Retained, Years
Challenger
Serves, Years
W 3 2
X 2 3
Y 1 4
Z 0 5
The respective AW values for defender retention and challenger use define the cash flows for
each option. Example 11.8 illustrates the procedure using the progressive example.
EXAMPLE 11.8 Keep or Replace the Kiln Case
PE
We have progressed to the point that the replacement study between the defender PT and
challenger GH was completed (Example 11.5). The defender was the clear choice with a
much smaller AW value ($8.50 M) than that of the challenger ($12.32 M). Now the
management of B&T is in a dilemma. They know the current tunnel kiln is much cheaper
than the new graphite hearth, but the prospect of future new business should not be dismissed.

11.5 Replacement Study over a Specified Study Period 311
TABLE 11–2 Replacement Study Options and Total AW Values, Example 11.8
Defender PT
Challenger GH
Option Years Retained AW, $ MYear Years Retained AW, $ MYear
A 1 8.50 5 14.21
B 2 9.06 4 15.08
C 3 9.59 3 16.31
D 4 10.49 2 18.21
E 5 11.16 1 22.10
F 6 11.47 0 —
The president asked, “Is it possible to determine when it is economically the cheapest to
purchase the new kiln, provided the current one is kept at least 1 year, but no more than 6
years, its remaining expected life?” The chief financial officer answered, yes, of course.
(a) Determine the answer for the president. (b) Discuss the next step in the analysis based on
the conclusion reached here.
Solution
(a) Actually, this is a quite easy question to answer, because all the information has been determined
previously. We know the MARR is 15% per year, the study period has been established
at 6 years, and the defender PT will stay in place between 1 and 6 years. Therefore,
the challenger GH will be considered for 0 to 5 years of service. The total AW values
were determined for the defender in Example 11.5 (Figure 11–6) and for the challenger in
Example 11.3 (Figure 11–3). They are repeated in Table 11–2 for convenience. Use the
procedure for a replacement study with a fixed study period.
Step 1: Succession options and AW values. There are six options in this case; the defender
is retained from 1 to 6 years while the challenger is installed from 0 to 5 years. We will
label them A through F. Figure 11–8 presents the options and the AW series for each option
from Table 11–2. No consideration of the fact that the challenger has an expected life of
12 years is made since the study period is fixed at 6 years.
Step 2: Selection of the best option. The PW value for each option is determined over the
6-year study period in column J of Figure 11–8. The conclusion is clearly to keep the
defender in place for 6 more years.
(b) Every replacement analysis has indicated that the defender should be retained for the near
future. If the analysis is to be carried further, the possibility of increased revenue based on
services of the challenger’s high-temperature and operating efficiency should be considered
next. In the introductory material, new business opportunities were mentioned. A revenue
increase for the challenger will reduce its AW of costs and possibly make it more
economically viable.
Conclusion: Keep defender all 6 years
Figure 11–8
PW values for 6-year study period replacement analysis, Example 11.8.

312 Chapter 11 Replacement and Retention Decisions
11.6 Replacement Value
Often it is helpful to know the minimum market value of the defender necessary to make the
challenger economically attractive. If a realizable market value or trade-in of at least this amount
can be obtained, from an economic perspective the challenger should be selected immediately.
This is a breakeven value between AW C and AW D ; it is referred to as the replacement value
(RV). Set up the relation AW C AW D with the market value for the defender identified as RV,
which is the unknown. The AW C is known, so RV can be determined. The selection guideline is
as follows:
If the actual market trade-in exceeds the breakeven replacement value , the challenger is the better
alternative and should replace the defender now.
Determination of the RV for a defender is an excellent opportunity to utilize the Goal Seek tool
in Excel. The target cell is the current market value, and the AW D value is forced to equal the
AW C amount. Example 11.9 discusses a replacement value of the progressive example.
EXAMPLE 11.9 Keep or Replace the Kiln Case
As one final consideration of the challenger kiln, you decide to determine what the trade-in
amount would have to be so that the challenger is the economic choice next year. This is based
on the ESL analysis that concluded the following (Examples 11.3 and 11.5):
Defender:
Challenger:
ESL n PT 1 year with AW PT $8.50 million
ESL n GH 12 years with AW GH $12.32 million
The original defender price was $25 million, and a current market value of $22 million was
estimated earlier (Figure 11–6). Since the installed kiln is known for retention of its market
value (MV), you are hopeful the difference between RV and estimated MV may not be so significant.
What will you discover RV to be? The MARR is 15% per year.
Solution
Set the AW relation for the defender for the ESL time of 1 year equal to AW GH $12.32
and solve for RV. The estimates for AOC and MV next year are in Figure 11–6; they are, in
$ million,
Year 1: AOC $−5.20 MV $22.0
12.32 RV(AP,15%,1) 22.00(AF,15%,1) 5.20
1.15RV 12.32 22.00 5.20
RV $25.32
Though the RV is larger than the defender’s estimated MV of $22 million, some flexibility in
the trade-in offer or the challenger’s first cost may cause the challenger to be economically
justifiable.
Comment
To find RV using a spreadsheet, return to Figure 11–6. In the Goal Seek template, the “set” cell
is the AW for 1 year (currently $8.50), and the required value is $12.32, the AW GH for its
ESL of 12 years. The “changing” cell is the current market value (cell F2), currently $22.00.
When “OK” is touched, $25.32 is displayed as the breakeven market value. This is the RV.
PE
CHAPTER SUMMARY
It is important in a replacement study to compare the best challenger with the defender. Best
(economic) challenger is described as the one with the lowest annual worth (AW) of costs for
some number of years . If the expected remaining life of the defender and the estimated life of the
challenger are specified, the AW values over these years are determined and the replacement
study proceeds. However, if reasonable estimates of the expected market value (MV) and AOC

for each year of ownership can be made, these year-by-year (marginal) costs help determine the
best challenger.
The economic service life (ESL) analysis is designed to determine the best challenger’s years
of service and the resulting lowest total AW of costs. The resulting n C and AW C values are used
in the replacement study. The same analysis can be performed for the ESL of the defender.
Replacement studies in which no study period (planning horizon) is specified utilize the annual
worth method of comparing two unequal-life alternatives. The better AW value determines
how long the defender is retained before replacement.
When a study period is specified for the replacement study, it is vital that the market value and
cost estimates for the defender be as accurate as possible. When the defender’s remaining life is
shorter than the study period, it is critical that the cost for continuing service be estimated carefully.
All the viable options for using the defender and challenger are enumerated, and their AW
equivalent cash flows are determined. For each option, the PW or AW value is used to select the
best option. This option determines how long the defender is retained before replacement.
Problems 313
PROBLEMS
Foundations of Replacement
11.1 In a replacement study, what is meant by “taking
the nonowner’s viewpoint”?
11.2 An asset that was purchased 3 years ago for
$100,000 is becoming obsolete faster than expected.
The company thought the asset would last
5 years and that its book value would decrease by
$20,000 each year and, therefore, be worthless at
the end of year 5. In considering a more versatile,
more reliable high-tech replacement, the company
discovered that the presently owned asset has a
market value of only $15,000. If the replacement is
purchased immediately at a first cost of $75,000
and if it will have a lower annual worth, what is the
amount of the sunk cost? Assume the company’s
MARR is 15% per year.
11.3 As a muscle car aficionado, a friend of yours likes
to restore cars of the 60s and 70s and sell them for
a profit. He started his latest project (a 1965 Shelby
GT350) four months ago and has a total of
$126,000 invested so far. Another opportunity has
come up (a 1969 Dodge Charger) that he is thinking
of buying because he believes he could sell it
for a profit of $60,000 after it is completely restored.
To do so, however, he would have to sell
the unfinished Shelby first. He thought that the
completely restored Shelby would be worth
$195,000, resulting in a tidy profit of $22,000, but
in its half-restored condition, the most he could get
now is $115,000. In discussing the situation with
you, he stated that if he could sell the Shelby now
and buy the Charger at a reduced price, he would
make up for the money he will lose in selling the
Shelby at a lower-than-desired price.
( a) What is wrong with this thinking?
( b) What is his sunk cost in the Shelby?
11.4 In conducting a replacement study wherein the
planning horizon is unspecified, list three assumptions
that are inherent in an annual worth analysis
of the defender and challenger.
11.5 A civil engineer who owns his own designbuild
operate company purchased a small crane 3 years
ago at a cost of $60,000. At that time, it was expected
to be used for 10 years and then traded in
for its salvage value of $10,000. Due to increased
construction activities, the company would prefer
to trade for a new, larger crane now that will cost
$80,000. The company estimates that the old crane
can be used, if necessary, for another 3 years, at
which time it would have a $23,000 estimated
market value. Its current market value is estimated
to be $39,000, and if it is used for another 3 years,
it will have M&O costs (exclusive of operator
costs) of $17,000 per year. Determine the values of
P , n , S , and AOC that should be used for the existing
crane in a replacement analysis.
11.6 Equipment that was purchased by Newport Corporation
for making pneumatic vibration isolators
cost $90,000 two years ago. It has a market value
that can be described by the relation $90,000
8000 k , where k is the years from time of purchase.
Experience with this type of equipment has shown
that the operating cost for the first 4 years is
$65,000 per year, after which it increases by $6300
per year. The asset’s salvage value was originally
estimated to be $7000 after a predicted 10-year
useful life. Determine the values of P , S , and AOC
if a replacement study is done ( a ) now and
( b ) 1 year from now.
11.7 A piece of equipment that was purchased 2 years
ago by Toshiba Imaging for $50,000 was expected
to have a useful life of 5 years with a $5000 salvage

314 Chapter 11 Replacement and Retention Decisions
value. Its performance was less than expected, and
it was upgraded for $20,000 one year ago. Increased
demand now requires that the equipment
be upgraded again for another $17,000 so that it
can be used for 3 more years. If upgraded, its annual
operating cost will be $27,000 and it will
have a $12,000 salvage after 3 years. Alternatively,
it can be replaced with new equipment priced at
$65,000 with operating costs of $14,000 per year
and a salvage value of $23,000 after 6 years. If replaced
now, the existing equipment will be sold for
$7000. Determine the values of P , S , AOC, and n
for the defender in a replacement study.
Economic Service Life
11.8 For equipment that has a first cost of $10,000 and
the estimated operating costs and year-end salvage
values shown below, determine the economic service
life at i 10% per year.
Year
Operating Cost,
$ per Year
Salvage
Value, $
1 1000 7000
2 1200 5000
3 1300 4500
4 2000 3000
5 3000 2000
11.9 To improve package tracking at a UPS transfer facility,
conveyor equipment was upgraded with
RFID sensors at a cost of $345,000. The operating
cost is expected to be $148,000 per year for the
first 3 years and $210,000 for the next 3 years. The
salvage value of the equipment is expected to be
$140,000 for the first 3 years, but due to obsolescence,
it won’t have a significant value after that.
At an interest rate of 10% per year, determine
(a) The economic service life of the equipment
and associated annual worth
( b) The percentage increase in the AW of cost if
the equipment is retained 2 years longer than
the ESL
11.10 Economic service life calculations for an asset are
shown below. If an interest rate of 10% per year
was used in making the calculations, determine the
values of P and S that were used in calculating the
AW for year 3.
Years
Retained
AW of First
Cost, $
AW of Operating
Cost, $ per Year
AW of
Salvage Value, $
1 51,700 15,000 35,000
2 27,091 17,000 13,810
3 18,899 19,000 6,648
4 14,827 21,000 4,309
5 12,398 23,000 2,457
11.11 The initial cost of a bridge that is expected to be in
place forever is $70 million. Maintenance can be
done at 1-, 2-, 3-, or 4-year intervals, but the longer
the interval between servicing, the higher the cost.
The costs of servicing are estimated at $83,000,
$91,000, $125,000, and $183,000 for intervals of 1
through 4 years, respectively. What interval should
be scheduled for maintenance to minimize the
overall equivalent annual cost? The interest rate is
8% per year.
11.12 A construction company bought a 180,000 metric
ton earth sifter at a cost of $65,000. The company
expects to keep the equipment a maximum of
7 years. The operating cost is expected to follow the
series described by 40,000 10,000 k , where k is
the number of years since it was purchased ( k 1,
2, . . . , 7). The salvage value is estimated to be
$30,000 for years 1 and 2 and $20,000 for years 3
through 7. At an interest rate of 10% per year, determine
the economic service life and the associated
equivalent annual cost of the sifter.
11.13 An engineer determined the ESL of a new
$80,000 piece of equipment and recorded the
calculations shown below. [Note that the numbers
are annual worth values associated with
various years of retention; that is, if the equipment
is kept for, say, 3 years, the AW (years 1
through 3) of the first cost is $32,169, the AW of
the operating cost is $51,000, and the AW of the
salvage value is $6042.] The engineer forgot to
enter the AW of the salvage value for 2 years of
retention. From the information available, determine
the following:
(a) The interest rate used in the ESL calculations.
( b) The salvage value after 2 years, if the total
AW of the equipment in year 2 was
$78,762. Use the interest rate determined
in part ( a ).
Years
Retained
AW of
First Cost, $
AW of
Operating Cost, $
AW of
Salvage Value, $
1 88,000 45,000 50,000
2 46,095 46,000 ?
3 32,169 51,000 6,042
11.14 A large, standby electricity generator in a hospital
operating room has a first cost of $70,000 and may
be used for a maximum of 6 years. Its salvage
value, which decreases by 15% per year, is described
by the equation S 70,000(1 – 0.15) n ,
where n is the number of years after purchase. The
operating cost of the generator will be constant at
$75,000 per year. At an interest rate of 12% per
year, what are the economic service life and the
associated AW value?

Problems 315
11.15 A piece of equipment has a first cost of $150,000, a
maximum useful life of 7 years, and a market (salvage)
value described by the relation S 120,000
20,000 k , where k is the number of years since it was
purchased. The salvage value cannot go below zero.
The AOC series is estimated using AOC 60,000
10,000 k . The interest rate is 15% per year. Determine
the economic service life ( a ) by hand solution,
using regular AW computations, and ( b ) by spreadsheet,
using annual marginal cost estimates.
11.16 Determine the economic service life and corresponding
AW value for a machine that has the following
cash flow estimates. Use an interest rate of
14% per year and hand solution.
Year
Salvage
Value, $
Operating
Cost, $ per Year
0 100,000 —
1 75,000 –28,000
2 60,000 –31,000
3 50,000 –34,000
4 40,000 –34,000
5 25,000 –34,000
6 15,000 –45,000
7 0 –49,000
11.17 Use the annual marginal costs to find the economic
service life for Problem 11.16 on a spreadsheet. Assume
the salvage values are the best estimates of future
market value. Develop an Excel chart of annual
marginal costs (MC) and AW of MC over 7 years.
11.18 Keep or Replace the Kiln Case
PE
In Example 11.3 , the market value (salvage value)
series of the proposed $38 million replacement kiln
(GH) dropped to $25 million in only 1 year and then
retained 75% of the previous year’s market value
through the remainder of its 12-year expected life.
Based on the experience with the current kiln, and
the higher-temperature capability of the replacement,
the new kiln is actually expected to retain
only 50% of the previous year’s value starting in
year 5. Additionally, the heating element replacement
in year 6 will probably cost $4 million, not
$2 million. And finally, the maintenance costs will
be considerably higher as the kiln ages. Starting in
year 5, the AOC is expected to increase by 25% per
year, not 10% as predicted earlier. The Manager of
Critical Equipment is now very concerned that the
ESL will be significantly decreased from the
12 years calculated earlier ( Example 11.3 ).
( a) Determine the new ESL and associated AW
value.
( b) In percentage changes, estimate how much
these new cost estimates may affect the
minimum-cost life and AW of cost estimate.
Replacement Study
11.19 In a one-year-later analysis, what action should be
taken if ( a ) all estimates are still current and the
year is n D , ( b ) all estimates are still current and the
year is not n D , and ( c ) the estimates have changed?
11.20 Based on company records of similar equipment, a
consulting aerospace engineer at Aerospatiale estimated
AW values for a presently owned, highly
accurate steel rivet inserter.
If Retained This
Number of Years
AW Value,
$ per Year
1 62,000
2 51,000
3 49,000
4 53,000
5 70,000
A challenger has ESL 2 years and AW C
$48,000 per year. ( a ) If the consultant must recommend
a replaceretain decision today, should
the company keep the defender or purchase the
challenger? Why? The MARR is 15% per year.
( b ) When should the next replacement evaluation
take place?
11.21 In planning a plant expansion, MedImmune has an
economic decision to make—upgrade the existing
controlled-environment rooms or purchase new
ones. The presently owned ones were purchased 4
years ago for $250,000. They have a current “quick
sale” value of $20,000, but for an investment of
$100,000 now, they would be adequate for another
4 years, after which they would be sold for $40,000.
Alternatively, new controlled-environment rooms
could be purchased at a cost of $270,000. They are
expected to have a 10-year life with a $50,000 salvage
value at that time. Determine whether the
company should upgrade or replace. Use a MARR
of 20% per year.
11.22 Three years ago, Witt Gas Controls purchased
equipment for $80,000 that was expected to have a
useful life of 5 years with a $9000 salvage value.
Increased demand necessitated an upgrade costing
$30,000 one year ago. Technology changes now require
that the equipment be upgraded again for another
$25,000 so that it can be used for 3 more
years. Its annual operating cost will be $47,000, and
it will have a $22,000 salvage after 3 years. Alternatively,
it can be replaced with new equipment that
will cost $68,000 with operating costs of $35,000
per year and a salvage value of $21,000 after
3 years. If replaced now, the existing equipment will
be sold for $9000. Calculate the annual worth of the
defender at an interest rate of 10% per year.

316 Chapter 11 Replacement and Retention Decisions
11.23 A recent environmental engineering graduate is trying
to decide whether he should keep his presently
owned car or purchase a more environmentally
friendly hybrid. A new car will cost $26,000 and have
annual operation and maintenance costs of $1200 per
year with an $8000 salvage value in 5 years (which is
its estimated economic service life).
The presently owned car has a resale value now
of $5000; one year from now it will be $3000, two
years from now $2500, and 3 years from now
$2200. Its operating cost is expected to be $1900
this year, with costs increasing by $200 per year.
The presently owned car will definitely not be kept
longer than 3 more years. Assuming used cars like
the one presently owned will always be available,
should the presently owned car be sold now, 1 year
from now, 2 years from now, or 3 years from now?
Use annual worth calculations at i 10% per year
and show your work.
11.24 A pulp and paper company is evaluating whether it
should retain the current bleaching process that
uses chlorine dioxide or replace it with a proprietary
“oxypure” process. The relevant information
for each process is shown. Use an interest rate of
15% per year to perform the replacement study.
Current Process
Oxypure Process
Original cost 6 450,000 —
years ago, $
Investment cost
— 700,000
now, $
Current market
25,000 —
value, $
Annual operating 180,000 70,000
cost, $year
Salvage value, $ 0 50,000
Remaining life,
years
5 10
11.25 A machine that is critical to the Phelps-Dodge copper
refining operation was purchased 7 years ago
for $160,000. Last year a replacement study was
performed with the decision to retain it for 3 more
years. The situation has changed. The equipment is
estimated to have a value of $8000 if “scavenged”
for parts now or anytime in the future. If kept in
service, it can be minimally upgraded at a cost of
$43,000, which will make it usable for up to
2 more years. Its operating cost is expected to be
$22,000 the first year and $25,000 the second year.
Alternatively, the company can purchase a new
system that will have an equivalent annual worth
of $47,063 per year over its ESL. The company
uses a MARR of 10% per year. Calculate the relevant
annual worth values, and determine when the
company should replace the machine.
11.26 A crushing machine that is a basic component of a
metal recycling operation is wearing out faster than
expected. The machine was purchased 2 years ago
for $400,000. At that time, the buyer thought the
machine would serve its needs for at least 5 years,
at which time the machine would be sold to a
smaller independent recycler for $80,000. Now,
however, the company thinks the market value of
the diminished machine is only $50,000. If it is
kept, the machine’s operating cost will be $37,000
per year for the next 2 years, after which it will be
scrapped for $1000. If it is kept for only 1 year, the
market value is estimated to be $10,000. Alternatively,
the company can outsource the process now
for a fixed cost of $56,000 per year. At an interest
rate of 10% per year, should the company continue
with the current machine or outsource the process?
11.27 The data associated with operating and maintaining
an asset are shown below. The company manager
has already decided to keep the machine for 1
more year (i.e., until the end of year 1), but you
have been asked to determine the cost of keeping it
1 more year after that. At an interest rate of 10%
per year, estimate the AW of keeping the machine
from year 1 to year 2.
Year Market Value, $
Operating Cost,
$ per Year
0 30,000 15,000
1 25,000 15,000
2 14,000 15,000
3 10,000 15,000
11.28 A machine that cost $120,000 three years ago can
be sold now for $54,000. Its market value for the
next 2 years is expected to be $40,000 and $20,000
one year and 2 years from now, respectively. Its
operating cost was $18,000 for the first 3 years of
its life, but the M&O cost is expected to be
$23,000 for the next 2 years. A new improved machine
that can be purchased for $138,000 will
have an economic life of 5 years, an operating
cost of $9000 per year, and a salvage value of
$32,000 after 5 years. At an interest rate of 10%
per year, determine if the presently owned machine
should be replaced now, 1 year from now, or
2 years from now.
11.29 The projected market value and M&O costs associated
with a presently owned machine are shown
(next page). An outside vendor of services has offered
to provide the service of the existing machine
at a fixed price per year. If the presently
owned machine is replaced now, the cost of the
fixed-price contract will be $33,000 for each of the
next 3 years. If the presently owned machine is

Problems 317
replaced next year or the year after that, the contract
price will be $35,000 per year. Determine if
and when the defender should be replaced with the
outside vendor using an interest rate of 10% per
year. Assume used equipment similar to the defender
will always be available.
Year Market Value, $
M&O Cost,
$ per Year
0 30,000 —
1 25,000 24,000
2 14,000 25,000
3 10,000 26,000
4 8,000 —
11.30 BioHealth, a biodevice systems leasing company,
is considering a new equipment purchase to replace
a currently owned asset that was purchased
2 years ago for $250,000. It is appraised at a current
market value of only $50,000. An upgrade is
possible for $200,000 now that would be adequate
for another 3 years of lease rights, after which the
entire system could be sold on the international
circuit for an estimated $40,000. The challenger,
which can be purchased for $300,000, has an expected
life of 10 years and a $50,000 salvage
value. Determine whether the company should upgrade
or replace at a MARR of 12% per year. Assume
the AOC estimates are the same for both
alternatives.
11.31 For the estimates in Problem 11.30, use a
spreadsheet-based analysis to determine the first
cost for the augmentation of the current system
that will make the defender and challenger break
even. Is this a maximum or minimum for the upgrade,
if the current system is to be retained?
11.32 Herald Richter and Associates, 5 years ago, purchased
for $45,000 a microwave signal graphical
plotter for corrosion detection in concrete structures.
It is expected to have the market values and
annual operating costs shown below for its remaining
useful life of up to 3 years. It could be
traded now at an appraised market value of $8000.
Year
Market Value
at End of Year, $
AOC,
$ per Year
1 6000 –50,000
2 4000 –53,000
3 1000 –60,000
A replacement plotter with new Internet-based,
digital technology costing $125,000 has an estimated
$10,000 salvage value after its 5-year life
and an AOC of $31,000 per year. At an interest rate
of 15% per year, determine how many more years
Richter should retain the present plotter. Solve
( a ) by hand and ( b ) by using a spreadsheet.
11.33 In the opportunity cost approach to replacement
analysis, what does the opportunity refer to?
11.34 State what is meant by the cash fl ow approach to
replacement analysis, and list two reasons why it is
not a good idea to use this method.
Replacement Study over a Specified Study Period
11.35 ABB Communications is considering replacing
equipment that had a first cost of $300,000 five
years ago. The company CEO wants to know if the
equipment should be replaced now or at any other
time over the next 3 years to minimize the cost of
producing miniature background suppression sensors.
Since the present equipment or the proposed
equipment can be used for any or all of the 3-year
period, one of the company’s industrial engineers
produced AW cost information for the defender
and challenger as shown below. The values represent
the annual costs of the respective equipment if
used for the indicated number of years. Determine
when the defender should be replaced to minimize
the cost to ABB for the 3-year study period using
an interest rate of 10% per year.
Number of
Years Kept
AW If Kept Stated
Number of Years, $ per Year
Defender
Challenger
1 22,000 29,000
2 24,000 26,000
3 27,000 25,000
11.36 The table below shows present worth calculations
of the costs associated with using a presently
owned machine (defender) and a possible replacement
(challenger) for different numbers of years.
Determine when the defender should be replaced
using an interest rate of 10% per year and a 5-year
study period. Show solutions ( a ) by hand and
( b ) by spreadsheet.
Number of Years
KeptUsed
PW If KeptUsed Stated
Number of Years, $
Defender
Challenger
1 36,000 89,000
2 75,000 96,000
3 125,000 102,000
4 166,000 113,000
5 217,000 149,000
11.37 Nano Technologies intends to use the newest and
finest equipment in its labs. Accordingly, a senior
engineer has recommended that a 2-year-old piece

318 Chapter 11 Replacement and Retention Decisions
of precision measurement equipment be replaced
immediately. This engineer believes it can be
shown that the proposed equipment is economically
advantageous at a 10% per year return and a
planning horizon of 3 years.
(a) Perform the replacement analysis using the
annual worth method for a specified 3-year
study period.
(b) Determine the challenger’s capital recovery
amount for the 3-year study period and the
expected full life. Comment on the effect
made by the 3-year study period.
Current
Proposed
Original purchase price, $ 30,000 40,000
Current market value, $ 17,000 —
Remaining life, years 5 15
Estimated value in 3 years, $ 9,000 20,000
Salvage value after 15 years, $ — 5,000
Annual operating cost, $ per year 8,000 3,000
11.38 An industrial engineer at a fiber-optic manufacturing
company is considering two robots to reduce
costs in a production line. Robot X will have a first
cost of $82,000, an annual maintenance and operation
(M&O) cost of $30,000, and salvage values of
$50,000, $42,000, and $35,000 after 1, 2, and
3 years, respectively. Robot Y will have a first cost
of $97,000, an annual M&O cost of $27,000, and
salvage values of $60,000, $51,000, and $42,000
after 1, 2, and 3 years, respectively. Which robot
should be selected if a 2-year study period is specified
at an interest rate of 15% per year and replacement
after 1 year is not an option?
11.39 A 3-year-old machine purchased for $140,000 is not
able to meet today’s market demands. The machine
can be upgraded now for $70,000 or sold to a subcontracting
company for $40,000. The current machine
will have an annual operating cost of $85,000
per year and a $30,000 salvage value in 3 years. If
upgraded, the presently owned machine will be retained
for only 3 more years, then replaced with a
machine to be used in the manufacture of several
other product lines. The replacement machine,
which will serve the company now and for at least
8 years, will cost $220,000. Its salvage value will be
$50,000 for years 1 through 4; $20,000 after 5 years;
and $10,000 thereafter. It will have an estimated operating
cost of $65,000 per year. You want to perform
an economic analysis at 15% per year using a
3-year planning horizon.
(a) Should the company replace the presently
owned machine now, or do it 3 years from now?
(b) Compare the capital recovery requirements for
the replacement machine (challenger) over the
study period and an expected life of 8 years.
11.40 Two processes can be used for producing a polymer
that reduces friction loss in engines. Process
K, which is currently in place, has a market value
of $165,000 now, an operating cost of $69,000 per
year, and a salvage value of $50,000 after 1 more
year and $40,000 after its maximum 2-year remaining
life. Process L, the challenger, will have a
first cost of $230,000, an operating cost of $65,000
per year, and salvage values of $100,000 after 1
year, $70,000 after 2 years, $45,000 after 3 years,
and $26,000 after its maximum expected 4-year
life. The company’s MARR is 12% per year. You
have been asked to determine which process to select
when ( a ) a 2-year study period is used and
( b ) a 3-year study period is used.
11.41 Keep or Replace the Kiln Case
In Example 11.8 , the in-place kiln and replacement
kiln (GH) were evaluated using a fixed study period
of 6 years. This is a significantly shortened
period compared to the expected 12-year life of
the challenger. Use the best estimates available
throughout this case to determine the impact on the
capital recovery amount for the GH kiln of shortening
the evaluation time from 12 to 6 years.
11.42 Nabisco Bakers currently employs staff to operate
the equipment used to sterilize much of the mixing,
baking, and packaging facilities in a large cookie
and cracker manufacturing plant in Iowa. The plant
manager, who is dedicated to cutting costs but not
sacrificing quality and hygiene, has the projected
data shown in the table below if the current system
were retained for up to its maximum expected life
of 5 years. A contract company has proposed a
turnkey sanitation system for $5.0 million per year
if Nabisco signs on for 4 to 10 years, and $5.5 million
per year for a shorter number of years.
Years Retained AW, $ per Year Close-Down Expense, $
0 — 3,000,000
1 2,300,000 2,500,000
2 2,300,000 2,000,000
3 3,000,000 1,000,000
4 3,000,000 1,000,000
5 3,500,000 500,000
(a)
PE
At a MARR 8% per year, perform a replacement
study for the plant manager with a
fixed study period of 5 years, when it is anticipated
that the plant will be shut down due
to the age of the facility and projected technological
obsolescence. As you perform the
study, take into account that regardless of the
number of years that the current sanitation
system is retained, a one-time close-down

Additional Problems and FE Exam Review Questions 319
cost will be incurred for personnel and equipment
during the last year of operation. ( Hint:
Calculate AW values for all combinations of
defenderchallenger options.)
( b) What is the percentage change in the AW
amount each year for the 5-year period? If
the decision to retain the current sanitation
system for all 5 years is made, what is the
economic disadvantage in AW compared to
that of the most economic retention period?
Replacement Value
11.43 In 2008, Amphenol Industrial purchased a new
quality inspection system for $550,000. The estimated
salvage value was $50,000 after 8 years.
Currently the expected remaining life is 3 years
with an AOC of $27,000 per year and an estimated
salvage value of $30,000. The new president has
recommended early replacement of the system
with one that costs $400,000 and has a 5-year
economic service life, a $45,000 salvage value,
and an estimated AOC of $50,000 per year. If the
MARR for the corporation is 12% per year, find
the minimum trade-in value now necessary to
make the president’s replacement economically
advantageous.
11.44 A CNC milling machine purchased by Proto Tool
and Die 10 years ago for $75,000 can be used for 3
more years. Estimates are an annual operating cost
of $63,000 and a salvage value of $25,000. A challenger
will cost $130,000 with an economic life of
6 years and an operating cost of $32,000 per year.
Its salvage value will be $45,000. On the basis of
these estimates, what market value for the existing
asset will render the challenger equally attractive?
Use an interest rate of 12% per year.
11.45 Hydrochloric acid, which fumes at room temperatures,
creates a very corrosive work environment,
causing steel tools to rust and equipment to fail
prematurely. A distillation system purchased 4
years ago for $80,000 can be used for only 2 more
years, at which time it will be scrapped with no
value. Its operating cost is $75,000 per year. A
more corrosion-resistant challenger will cost
$220,000 with an operating cost of $49,000 per
year. It is expected to have a $30,000 salvage
value after its 6-year ESL. At an interest rate of
10% per year, what minimum replacement value
now for the present system will render the challenger
attractive?
11.46 Engine oil purifier machines can effectively remove
acid, pitch, particles, water, and gas from
used oil. Purifier A was purchased 5 years ago for
$90,000. Its operating cost is higher than expected,
so if it is not replaced now, it will likely be used for
only 3 more years. Its operating cost this year will
be $140,000, increasing by $2000 per year through
the end of its useful life, at which time it will be
donated for its recyclable scrap value. A more efficient
challenger, purifier B, will cost $150,000
with a $50,000 salvage value after its 8-year ESL.
Its operating cost is expected to be $82,000 for
year 1, increasing by $500 per year thereafter.
What is the market value for A that will make the
two purifiers equally attractive at an interest rate of
12% per year?
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
11.47 In conducting a replacement study, all of the following
are correct viewpoints for the analyst except:
( a) Consultant’s
( b) Owner’s
( c) Outsider’s
( d) Nonowner’s
11.48 A sunk cost is the difference between:
( a) The first cost and the salvage value
( b) The market value and the salvage value
( c) The first cost and the market value
( d) The book value and the market value
11.49 A truck was purchased 3 years ago for $45,000 and
can be sold today for $24,000. The operating costs
are $9000 per year, and it is expected to last 4 more
years with a $5000 salvage value. A new truck,
which will perform that same service, can be purchased
for $50,000, and it will have a life of
10 years with operating costs of $28,000 per year
and a $10,000 salvage value. The value that should
be used as P for the presently owned vehicle in a
replacement study is:
(a) $45,000
(b) $5000
(c) $50,000
(d) $24,000
11.50 The economic service life of an asset with the PW
and AW values on the next page is:
(a) 1 year
(b) 2 years
(c) 3 years
(d) 4 years

320 Chapter 11 Replacement and Retention Decisions
Years n
Present Worth If
Kept n Years, $
Annual Worth If
Kept n Years,
$Year
1 15,455 17,000
2 28,646 16,286
3 45,019 18,102
4 57,655 18,188
5 72,867 19,222
11.51 In looking for ways to cut costs and increase
profit (to make the company’s stock go up), one
of the company’s industrial engineers (IEs) determined
that the equivalent annual worth of an
existing machine over its remaining useful life of
3 years will be $–70,000 per year. The IE also
determined that a replacement with more advanced
features will have an AW of $–80,000 per
year if it is kept for 2 years or less, $–75,000 if it
is kept between 3 and 4 years, and $–65,000 if it
is kept for 5 to 10 years. If the engineer uses a
3-year study period and an interest rate of 15%
per year, she should recommend that the existing
machine be:
(a) Replaced now
(b) Replaced 1 year from now
(c) Replaced 2 years from now
(d) Not replaced
11.52 The equivalent annual worth of an existing machine
at American Semiconductor is estimated to
be $–70,000 per year over its remaining useful life
of 3 years. It can be replaced now or later with a
machine that will have an AW of $–90,000 per
year if it is kept for 2 years or less, –$65,000 if it
is kept between 3 and 5 years, and $–110,000 if it
is kept for 6 to 8 years. The company wants an
analysis of what it should do for a 3-year study
period at an interest rate of 15% per year. The replacement
must be made now or 3 years from
now, according to the department supervisor. You
should recommend that the existing machine be
replaced:
(a) Now
(b) 1 year from now
(c) 2 years from now
(d) Can’t tell; must find AW of different retention
combinations for the two machines
11.53 The cost characteristics of a CO testing machine at
Dytran Instruments are shown below. The cost of a
new tester is $100,000. The equation for determining
the AW of keeping the tester for 2 years is:
(a) AW 100,000( AP,i,8) [42,000( PF,i,1)
47,000( PF,i ,2)](AP,i,8) 40,000( AF,i,8)
(b) AW 100,000( AP,i ,2) [42,000( PF,i,1)
47,000( PF,i,2)](AP,i ,2) 40,000( AF,i ,2)
( c) AW 100,000( AP,i, 2) 47,000
40,000( AF,i ,2)
(d) AW 100,000( AP,i, 2) 42,000
40,000( AF,i, 2)
Machine Age,
Years
M&O Costs,
$ per Year
Salvage Value
at End of Year, $
1 42,000 60,000
2 47,000 40,000
3 49,000 31,000
4 50,000 24,000
5 52,000 15,000
6 54,000 10,000
7 63,000 10,000
8 67,000 10,000
11.54 An engineer with Haliburton calculated the AW
values shown for a presently owned machine by
using estimates he obtained from the vendor and
company records.
Retention
Period, Years
AW Value,
$ per Year
1 92,000
2 81,000
3 87,000
4 89,000
5 95,000
A challenger has an economic service life of
7 years with an AW of $–86,000 per year. Assume
that used machines like the one presently owned
will always be available and that the MARR is
12% per year. If all future costs remain as estimated
for the analysis, the company should purchase
the challenger:
(a) Now
(b) After 2 years
(c) After 3 years
(d) Never
11.55 The annual worth values for a defender, which can
be replaced with a similar used asset, and a challenger
are estimated. The defender should be replaced:
(a) Now
(b) 1 year from now
(c) 2 years from now
(d) 3 years from now
Number of
Years Retained
AW Value, $ per Year
Defender
Challenger
1 14,000 21,000
2 13,700 18,000
3 16,900 13,100
4 17,000 15,600
5 18,000 17,500

Case Study 321
CASE STUDY
WILL THE CORRECT ESL PLEASE STAND?
Background
New pumper system equipment is under consideration by a
Gulf Coast chemical processing plant. One crucial pump
moves highly corrosive liquids from specially lined tanks on
intercoastal barges into storage and preliminary refining facilities
dockside. Because of the variable quality of the raw
chemical and the high pressures imposed on the pump chassis
and impellers, a close log is maintained on the number of
hours per year that the pump operates. Safety records and
pump component deterioration are considered critical control
points for this system. As currently planned, rebuild and
M&O cost estimates are increased accordingly when cumulative
operating time reaches the 6000-hour mark.
Information
You are the safety engineer at the plant. Estimates made for
this pump are as follows:
First cost:
Rebuild cost:
$800,000
$150,000 whenever 6000 cumulative
hours are logged. Each
rework will cost 20% more than
the previous one. A maximum of
3 rebuilds is allowed.
M&O costs: $25,000 for each year 1 through 4
$40,000 per year starting the
year after the first rebuild, plus
15% per year thereafter
MARR: 10% per year
Based on previous logbook data, the current estimates for
number of operating hours per year are as follows:
Year
Hours per Year
1 500
2 1500
3 on 2000
Case Study Questions
1. Determine the economic service life of the pump. How
does the ESL compare with the maximum allowed
rebuilds?
2. The plant superintendent told you, the safety engineer,
that only one rebuild should be planned for, because
these types of pumps usually have their minimum-cost
life before the second rebuild. Determine a market
value for this pump that will force the ESL to be 6 years.
Comment on the practicality of ESL 6 years, given
the MV calculated.
3. In a separate conversation, the line manager told you
to not plan for a rebuild after 6000 hours, because
the pump will be replaced after a total of 10,000
hours of operation. The line manager wants to know
what the base AOC in year 1 can be to make the ESL
6 years. He also told you to assume now that the 15%
growth rate applies from year 1 forward. How does
this base AOC value compare with the rebuild cost
after 6000 hours?
4. What do you think of these suggestions from the plant
superintendent and the line manager?

CHAPTER 12
Independent
Projects with
Budget
Limitation
LEARNING OUTCOMES
Purpose: Select independent projects for funding when there is a limitation on the amount of capital available for investment.
SECTION TOPIC LEARNING OUTCOME
12.1 Capital rationing • Explain how a capital budgeting problem is
approached.
12.2 Equal-life projects • Use PW-based capital budgeting to select from
several equal-life independent projects.
12.3 Unequal-life projects • Use PW-based capital budgeting to select from
several unequal-life independent projects.
12.4 Linear programming • Set up the linear programming model and use the
Solver spreadsheet tool to select projects.
12.5 Ranking options • Use the internal rate of return (IROR) and
profitability index (PI) to rank and select from
independent projects.

I
n most of the previous economic comparisons, the alternatives have been mutually
exclusive; only one could be selected. If the projects are not mutually exclusive,
they are categorized as independent of one another, as discussed at the beginning
of Chapter 5. Now we learn techniques to select from several independent projects. It
is possible to select any number of projects from none (do nothing) to all viable projects.
There is virtually always some upper limit on the amount of capital available for investment
in new projects. This limit is considered as each independent project is economically
evaluated. The techniques applied are called capital budgeting methods, also referred to
as capital rationing. They determine the economically best rationing of initial investment
capital among independent projects based upon different measures, such as PW, ROR, and
the profitability index. These three are discussed here.
12.1 An Overview of Capital Rationing
among Projects
Investment capital is a scarce resource for all corporations; thus there is virtually always a limited
amount to be distributed among competing investment opportunities. When a corporation has
several options for placing investment capital, a “reject or accept” decision must be made for
each project. Effectively, each option is independent of other options, so the evaluation is performed
on a project-by-project basis. Selection of one project does not impact the selection decision
for any other project. This is the fundamental difference between mutually exclusive alternatives
and independent projects.
The term project is used to identify each independent option. We use the term bundle to
identify a collection of independent projects. The term mutually exclusive alternative continues
to identify a project when only one may be selected from several.
There are two exceptions to purely independent projects: A contingent project is one that has
a condition placed upon its acceptance or rejection. Two examples of contingent projects A and
B are as follows: A cannot be accepted unless B is accepted; and A can be accepted in lieu of B,
but both are not needed. A dependent project is one that must be accepted or rejected based on
the decision about another project(s). For example, B must be accepted if both A and C are accepted.
In practice, these complicating conditions can be bypassed by forming packages of related
projects that are economically evaluated themselves as independent projects along with the
remaining, unconditioned projects.
A capital budgeting study has the following characteristics:
• Several independent projects are identified, and net cash flow estimates are available.
• Each project is either selected entirely or not selected; that is, partial investment in a project
is not possible.
• A stated budgetary constraint restricts the total amount available for investment. Budget
constraints may be present for the first year only or for several years. This investment limit
is identified by the symbol b .
• The objective is to maximize the return on the investments using a measure of worth, such
as the PW value.
Independent p roject
selection
Limited budget
By nature, independent projects are usually quite different from one another. For example, in the
public sector, a city government may develop several projects to choose from: drainage, city
park, metro rail, and an upgraded public bus system. In the private sector, sample projects may
be a new warehousing facility, expanded product base, improved quality program, upgraded information
system, automation, and acquisition of another firm. The size of the study can range
from only four or five projects to a complex study involving 50 to 100 projects. The typical
capital budgeting problem is illustrated in Figure 12–1. For each independent project there is an
initial investment, project life, and estimated net cash flows that can include a salvage value.
Present worth analysis using the capital budgeting process is the recommended method to
select projects. The general selection guideline is as follows:
Accept projects with the best PW values determined at the MARR over the project life, provided
the investment capital limit is not exceeded.

324 Chapter 12 Independent Projects with Budget Limitation
Figure 12–1
Basic characteristics of a
capital budgeting study.
Independent
projects
Initial
investment
Estimates
Estimated
net cash flows
A.
Life
Capital
investment
limit
Investment
B.
Life
(can’t invest
more than
this much)
C.
Life
Select 0 to all 3 projects
Objective: Maximize PW
value of selection
within capital limit
Equal-service requirement
This guideline is not different from that used for selection in previous chapters for independent
projects. As before, each project is compared with the do-nothing project; that is, incremental
analysis between projects is not necessary. The primary difference now is that the amount of
money available to invest is limited, thus the title capital budgeting or rationing . A specific solution
procedure that incorporates this constraint is needed.
Previously, PW analysis had the requirement of equal service between alternatives. This assumption
is not necessary for capital rationing, because there is no life cycle of a project beyond
its estimated life. Rather, the selection guideline has the following implied assumption.
When the present worth at the MARR over the respective project life is used to select projects,
the reinvestment assumption is that all positive net cash flows are reinvested at the MARR from
the time they are realized until the end of the longest-lived project.
Opportunity cost
This fundamental assumption is demonstrated to be correct at the end of Section 12.3, which
treats PW-based capital rationing for unequal-life projects.
Another dilemma of capital rationing among independent projects concerns the flexibility of the
capital investment limit b . The limit may marginally disallow an acceptable project that is next in
line for acceptance. For example, assume project A has a positive PW value at the MARR. If A will
cause the capital limit of $5,000,000 to be exceeded by only $1000, should A be included in the PW
analysis? Commonly, a capital investment limit is somewhat flexible, so project A would usually be
included. However, in the examples here, we will not exceed a stated investment limit.
As we learned earlier (Sections 1.9 and 10.1), the rate of return on the first unfunded project
is an opportunity cost . The lack of capital to fund the next project defines the ROR level that
is forgone. The opportunity cost will vary with each set of independent projects evaluated, but
over time it provides information to fine-tune the MARR and other measures used by the
company in future evaluations.

12.2 Capital Rationing Using PW Analysis of Equal-Life Projects 325
It is common to rank independent projects and select from them based on measures other than
PW at the MARR. Two are the internal rate of return (IROR), discussed in Chapter 7, and the
profi tability index (PI) , also called the present worth index (PWI), introduced in Chapter 9. Neither
of these measures guarantees an optimal PW-based selection. The capital budgeting process,
covered in the next 3 sections, does find the optimal solution for PW values. We recommend use
of this PW-based technique; however, it should be recognized that both PI and IROR usually
provide excellent, near-optimal selections, and they both work very well when the number of
projects is large. Application of these two measures is presented in Section 12.5.
12.2 Capital Rationing Using PW Analysis
of Equal-Life Projects
To select from projects that have the same expected life while investing no more than the limit b ,
first formulate all mutually exclusive bundles —one project at a time, two at a time, etc. Each
feasible bundle must have a total investment that does not exceed b . One of these bundles is the
do-nothing (DN) project. The total number of bundles for m projects is 2 m . The number increases
rapidly with m . For m 4, there are 2 4 16 bundles, and for m 16, 2 16 65,536 bundles.
Then the PW of each bundle is determined at the MARR. The bundle with the largest PW value
is selected.
To illustrate the development of mutually exclusive bundles, consider these four projects with
equal lives.
Project Initial Investment, $
A 10,000
B 5,000
C 8,000
D 15,000
If the investment limit is b $25,000, of the 16 bundles, there are 12 feasible ones to evaluate.
The bundles ABD, ACD, BCD, and ABCD have investment totals that exceed $25,000. The
viable bundles are shown below.
Projects
Total Initial
Investment, $
Projects
Total Initial
Investment, $
A 10,000 AD 25,000
B 5,000 BC 13,000
C 8,000 BD 20,000
D 15,000 CD 23,000
AB 15,000 ABC 23,000
AC 18,000 Do nothing 0
The procedure to conduct a capital budgeting study using PW analysis is as follows:
1. Develop all mutually exclusive bundles with a total initial investment that does not exceed
the capital limit b .
2. Sum the net cash flows NCF jt for all projects in each bundle j ( j 1, 2, . . . , 2 m ) and each
year t ( t 1, 2, . . . , n j ). Refer to the initial investment of bundle j at time t 0 as NCF j 0 .
3. Compute the present worth value PW j for each bundle at the MARR.
PW j PW of bundle net cash flows initial investment
tn j
NCF jt (PF,i,t) NCF j0 [12.1]
t1
4. Select the bundle with the (numerically) largest PW j value.
Selecting the maximum PW j means that this bundle has a PW value larger than any other bundle.
Any bundle with PW j 0 is discarded, because it does not produce a return of at least the MARR.

326 Chapter 12 Independent Projects with Budget Limitation
EXAMPLE 12.1
The projects review committee of Microsoft has $20 million to allocate next year to new software
product development. Any or all of five projects in Table 12–1 may be accepted. All
amounts are in $1000 units. Each project has an expected life of 9 years. Select the project(s)
if a 15% return is expected.
TABLE 12–1 Five Equal-Life Independent Projects ($1000 Units)
Project
Initial
Investment, $
Annual
Net Cash Flow, $
Project
Life, Years
A 10,000 2870 9
B 15,000 2930 9
C 8,000 2680 9
D 6,000 2540 9
E 21,000 9500 9
Solution
Use the procedure above with b $20,000 to select one bundle that maximizes present worth.
Remember the units are in $1000.
5
1. There are 2 32 possible bundles. The eight bundles that require no more than $20,000
in initial investments are described in columns 2 and 3 of Table 12–2. The $21,000 investment
for E eliminates it from all bundles.
TABLE 12–2 Summary of Present Worth Analysis of Equal-Life Independent
Projects ($1000 Units)
Bundle
j
(1)
Projects
Included
(2)
Initial
Investment
NCF j0 , $
(3)
Annual Net
Cash Flow
NCF j , $
(4)
Present
Worth
PW j , $
(5)
1 A 10,000 2,870 3,694
2 B 15,000 2,930 1,019
3 C 8,000 2,680 4,788
4 D 6,000 2,540 6,120
5 AC 18,000 5,550 8,482
6 AD 16,000 5,410 9,814
7 CD 14,000 5,220 10,908
8 Do nothing 0 0 0
2. The bundle net cash flows, column 4, are the sum of individual project net cash
flows.
3. Use Equation [12.1] to compute the present worth for each bundle. Since the annual NCF
and life estimates are the same for a bundle, PW j reduces to
PW j NCF j ( P A ,15%,9) NCF j 0
4. Column 5 of Table 12–2 summarizes the PW j values at i 15%. Bundle 2 does not return
15%, since PW 2 0. The largest is PW 7 $10,908; therefore, invest $14 million in C and
D. This leaves $6 million uncommitted.
This analysis assumes that the $6 million not used in this initial investment will return the
MARR by placing it in some other, unspecified investment opportunity.

12.3 Capital Rationing Using PW Analysis of Unequal-Life Projects 327
12.3 Capital Rationing Using PW Analysis
of Unequal-Life Projects
Usually independent projects do not have the same expected life. As stated in Section 12.1, the
PW method for solution of the capital budgeting problem assumes that each project will last for
the period of the longest-lived project n L . Additionally, reinvestment of any positive net cash
flows is assumed to be at the MARR from the time they are realized until the end of the longestlived
project, that is, from year n j through year n L . Therefore, use of the LCM of lives is not
necessary , and it is correct to use Equation [12.1] to select bundles of unequal-life projects by
PW analysis using the procedure of the previous section.
EXAMPLE 12.2
For MARR 15% per year and b $20,000, select from the following independent projects.
Solve by hand and by spreadsheet.
Project
Initial
Investment, $
Annual Net
Cash Flow, $
Project
Life, Years
A 8,000 3870 6
B 15,000 2930 9
C 8,000 2680 5
D 8,000 2540 4
Solution by Hand
The unequal-life values make the net cash flows vary over a bundle’s life, but the selection
procedure is the same as above. Of the 2 4 16 bundles, 8 are economically feasible. Their PW
values by Equation [12.1] are summarized in Table 12–3. As an illustration, for bundle 7:
PW 7 16,000 5220( P A ,15%,4) 2680( P F ,15%,5) $235
Select bundle 5 (projects A and C) for a $16,000 investment.
TABLE 12–3 Present Worth Analysis for Unequal-Life Independent Projects,
Example 12.2
Bundle J
(1)
Project
(2)
Initial
Investment,
NCF j0 , $
(3)
Net Cash Flows
Year t
(4)
NCF jt , $
(5)
Present
Worth
PW j , $
(6)
1 A 8,000 1–6 3,870 6,646
2 B 15,000 1–9 2,930 1,019
3 C 8,000 1–5 2,680 984
4 D 8,000 1–4 2,540 748
5 AC 16,000 1–5 6,550 7,630
6 3,870
6 AD 16,000 1–4 6,410 5,898
5–6 3,870
7 CD 16,000 1–4 5,220 235
5 2,680
8 Do nothing 0 0 0
Solution by Spreadsheet
Figure 12–2 presents a spreadsheet with the same information as in Table 12–3. It is necessary
to initially develop the mutually exclusive bundles manually and total net cash flows each year
using each project’s NCF. The NPV function is used to determine PW for each bundle j over
its respective life. Bundle 5 (projects A and C) has the largest PW value (row 16).

328 Chapter 12 Independent Projects with Budget Limitation
D7E7
NPV($B$1,F7:F15) F6
Figure 12–2
Computation of PW values for independent project selection, Example 12.2.
The rest of this section will help you understand why solution of the capital budgeting problem
by PW evaluation using Equation [12.1] is correct. The following logic verifies the assumption of
reinvestment at the MARR for all net positive cash flows when project lives are unequal. Refer to
Figure 12–3, which uses the general layout of a two-project bundle. Assume each project has the
same net cash flow each year. The P A factor is used for PW computation. Define n L as the life of
the longer-lived project. At the end of the shorter-lived project, the bundle has a total future worth
of NCF j (F A ,MARR,n j ) as determined for each project. Now, assume reinvestment at the MARR
from year n j 1 through year n L (a total of n L − n j years). The assumption of the return at the MARR
is important; this PW approach does not necessarily select the correct projects if the return is not
at the MARR. The results are the two future worth arrows in year n L in Figure 12–3. Finally,
Figure 12–3
Representative cash flows
used to compute PW for a
bundle of two independent
unequal-life projects
by Equation [12.1].
PW B
FW B
Project B
Investment
for B
n B = n L
FW A
PW A
Future worth
Period of
reinvestment
at MARR
Project A
Investment
for A
n A
n L
Bundle PW = PW A +PW B

12.4 Capital Budgeting Problem Formulation Using Linear Programming 329
PW = $235
$5220
$2680
FW = $57,111
Figure 12–4
Initial investment and
cash flows for bundle 7,
projects C and D,
Example 17.2.
0
1 2 3 4 5 6 7 8 9
$16,000
compute the bundle PW value in the initial year. This is the bundle PW PW A PW B . In general
form, the bundle j present worth is
PW j NCF j ( F A ,MARR, n j )( F P ,MARR, n L n j )( P F ,MARR, n L ) [12.2]
Substitute the symbol i for the MARR, and use the factor formulas to simplify.
PW j NCF (1 i) n j
1
j
——————(1 i)
i
n L n j
———— 1
(1 i) n L
NCF j [ (1 i) n j
1
——————
i(1 i) n j ] [12.3]
NCF j (PA,i,n j )
Since the bracketed expression in Equation [12.3] is the ( P A , i , n j ) factor, computation of PW j for
n j years assumes reinvestment at the MARR of all positive net cash flows until the longest-lived
project is completed in year n L .
To demonstrate numerically, consider bundle j 7 in Example 12.2. The evaluation is in
Table 12–3, and the net cash flow is pictured in Figure 12–4. Calculate the future worth in year 9,
which is the life of the longest-lived project (B).
FW 5220( F A ,15%,4)( F P ,15%,5) 2680( F P ,15%,4) $57,111
The present worth at the initial investment time is
PW 16,000 57,111( P F ,15%,9) $235
The PW value is the same as PW 7 in Table 12–3 and Figure 12–2. This demonstrates the reinvestment
assumption for positive net cash flows. If this assumption is not realistic, the PW analysis
must be conducted using the LCM of all project lives.
12.4 Capital Budgeting Problem Formulation
Using Linear Programming
The procedure discussed above requires the development of mutually exclusive bundles one
project at a time, two projects at a time, etc., until all 2 m bundles are developed and each one is
compared with the capital limit b . As the number of independent projects increases, this process
becomes prohibitively cumbersome and unworkable. Fortunately, the capital budgeting problem
can be stated in the form of a linear programming model. The problem is formulated using the
integer linear programming (ILP) model, which means simply that all relations are linear and that
the variable x can take on only integer values. In this case, the variables can only take on the
values 0 or 1, which makes it a special case called the 0-or-1 ILP model. The formulation in
words follows.

330 Chapter 12 Independent Projects with Budget Limitation
Maximize: Sum of PW of net cash flows of independent projects.
Constraints:
• Capital investment constraint is that the sum of initial investments must not exceed a specified limit.
• Each project is completely selected or not selected.
For the math formulation, define b as the capital investment limit, and let x k ( k 1 to m projects)
be the variables to be determined. If x k 1, project k is completely selected; if x k 0, project k
is not selected. Note that the subscript k represents each independent project, not a mutually
exclusive bundle.
If the sum of PW of the net cash flows is Z , the math programming formulation is as follows:
Maximize:
Constraints:
km
k1
km
k1
PW k x k Z
NCF k 0 x k b
x k 0 or 1 for k 1, 2, . . . , m
[12.4]
EXAMPLE 12.3
The PW k of each project is calculated using Equation [12.1] at MARR i .
PW k PW of project net cash flows for n k years
tn k
NCF kt (PF,i,t) NCF k0 [12.5]
t1
Computer solution is accomplished by a linear programming software package which treats the ILP
model. Also, Excel and its optimizing tool Solver can be used to develop the formulation and select
the projects. The Solver tool is similar to Goal Seek with significantly more capabilities. For example,
Solver allows the target cell to be maximized, minimized, or set to a specific value. This means that the
function Z in Equation [12.4] can be maximized. Also, multiple changing cells can be identified, so
the 0 or 1 value of the unknowns can be determined. Additionally, with the added capability to include
constraints, the investment limit b and 0-or-1 requirement on the unknowns in Equation [12.4] can be
accommodated. Solver is explained in detail in Appendix A, and Example 12.3 illustrates its use.
Review Example 12.2. ( a ) Formulate the capital budgeting problem using the math programming
model presented in Equation [12.4]. ( b ) Select the projects using Solver.
Solution
(a) Define the subscript k 1 through 4 for the four projects, which are relabeled as 1, 2, 3,
and 4. The capital investment limit is b $20,000 in Equation [12.4].
Maximize:
Constraints:
k4
k1
k4
k1
PW k x k Z
NCF k0 x k 20,000
x k 0 or 1 for k 1, 2, 3, 4
Now, substitute the PW k and NCF k 0 values from Table 12–3 into the model. Plus signs are
used for all values in the budget constraint. We have the complete 0-or-1 ILP formulation.
Maximize:
6646 x 1 1019 x 2 984 x 3 748 x 4 Z
Constraints: 8000 x 1 15,000 x 2 8000 x 3 8000 x 4 20,000
x 1 , x 2 , x 3 , and x 4 0 or 1
The maximum PW is $7630, and the solution from Example 12.2 is written as
x 1 1 x 2 0 x 3 1 x 4 0

12.4 Capital Budgeting Problem Formulation Using Linear Programming 331
NPV($B$1,E7:E18) E6
Figure 12–5
Spreadsheet and Solver template configured to solve a capital budgeting problem, Example 12.3.
( b ) Figure 12–5 presents a spreadsheet template developed to select from six or fewer independent
projects with 12 years or less of net cash flow estimates per project. The spreadsheet
template can be expanded in either direction if needed. Figure 12–5 (inset) shows the
Solver parameters set to solve this example for four projects and an investment limit of
$20,000. The descriptions below and the cell tag identify the contents of the rows and cells
in Figure 12–5, and their linkage to Solver parameters.
Rows 4 and 5: Projects are identified by numbers to distinguish them from spreadsheet
column letters. Cell I5 is the expression for Z , the sum of the PW values for
the projects. This is the target cell for Solver to maximize.
Rows 6 to 18: These are initial investments and net cash flow estimates for each
project. Zero values that occur after the life of a project need not be entered;
however, any $0 estimates that occur during a project’s life must be entered.
Row 19: The entry in each cell is 1 for a selected project and 0 if not selected. These
are the changing cells for Solver. Since each entry must be 0 or 1, a binary constraint
is placed on all row 19 cells in Solver, as shown in Figure 12–5. When a
problem is to be solved, it is best to initialize the spreadsheet with 0s for all projects.
Solver will find the solution to maximize Z .
Row 20: The NPV function is used to find the PW for each net cash flow series. The
NPV functions are developed for any project with a life up to 12 years at the
MARR entered in cell B1.
Row 21: When a project is selected, the contribution to the Z function is shown by
multiplying rows 19 and 20.
Row 22: This row shows the initial investment for the selected projects. Cell I22 is
the total investment. This cell has the budget limitation placed on it by the constraint
in Solver. In this example, the constraint is I22 $20,000.
To solve the example, set all values in row 19 to 0, set up the Solver parameters as described
above, and click on Solve. (Since this is a linear model, the Solver options choice “Assume Linear
Model” may be checked, if desired.) If needed, further directions on saving the solution,
making changes, etc., are available in Appendix A, Section A.5, and on the Excel help function.
For this problem, the selection is projects 1 and 3 with Z $7630, the same as determined
previously, and $16,000 of the $20,000 limit is invested.

332 Chapter 12 Independent Projects with Budget Limitation
12.5 Additional Project Ranking Measures
The PW-based method of solving a capital budgeting problem covered in Sections 2.2 to 2.4
provides an optimal solution that maximizes the PW of the competing projects. However, it is
very common in industrial, professional, and government settings to learn that the rate of return
is the basis for ranking projects. The internal rate of return (IROR), as we learned in Section
7.2, is determined by setting a PW or AW relation equal to zero and solving for i * — the
IROR. Using a PW basis and the estimated net cash flow (NCF) series for each project j , solve
for i * in the relation
tn j
0 NCF jt (PF,i*,t) NCF j0 [12.6]
t1
This is the same as Equation [12.1] set equal to zero with i i * as the unknown. The spreadsheet
function RATE or IRR will provide the same answer. The selection guideline is as follows:
Independent project
selection
Once the project ranking by IROR is complete, select all projects in order without exceeding
the investment limit b.
If there is no budget limit, select all projects that have IROR MARR.
The ordering of projects using IROR ranking may differ from the PW-based ranking we used in
previous sections. This can occur because IROR ranking maximizes the overall rate of return, not
necessarily the PW value. The use of IROR ranking is illustrated in the next example.
Another common ranking method is the profitability index (PI) that we learned in
Section 9.2. This is a “bang for the buck” measure that provides a sense of getting the most for
the investment dollar over the life of the project. (Refer to Section 9.2 for more details.) When
utilized for project ranking, it is often called the present worth index (PWI); however, we will
use the PI term for consistency, and because there are other ways to mathematically define measures
also referred to as a PW index. The PI measure is defined as
tn j
NCF jt (PF, i, t)
———————————
PW of net cash flows
PW of initial investment t1
————————
NCF j0
[12.7]
Note that the denominator includes only the initial investment, and it is its absolute value that is
used. The numerator has only cash flows that result from the project for years 1 through its life
n j . Salvage value, if there is one estimated, is incorporated into the numerator. Similar to the
previous case, the selection guideline is as follows:
Independent project
selection
Once the project ranking by PI is complete, select all projects in order without exceeding the
investment limit b.
If there is no budget limit, select all projects that have PI 1.0.
Depending upon the project NCF estimates, the PI ranking can differ from the IROR ranking . Example
12.4 compares results using the different ranking methods—IROR, PI, and PW values. None
of these results are incorrect; they simply maximize different measures, as you will see. The use of
IROR, PI, or other measures is common when there are a large number of projects, because the PW
basis (when solved by hand, Solver, or ILP software) becomes increasingly cumbersome as the number
of possible mutually exclusive bundles grows using the formula 2 m . Additionally, greater complexity
is introduced when dependent and contingent projects are involved.
EXAMPLE 12.4
Georgia works as a financial analyst in the Management Science Group of General Electronics.
She has been asked to recommend which of the five projects detailed in Table 12–4 should be
funded if the MARR is 15% per year and the investment budget limit for next year is a firm

12.5 Additional Project Ranking Measures 333
TABLE 12–4 IROR, PI, and PW Values for Five Projects, Example 12.4
Projects 1 2 3 4 5
Investment, NCF 0 , $1000 8,000 15,000 8,000 8,000 5,000
NCF, $1000 per year 4,000 2,900 2,700 2,500 2,600
Life n j , years 6 9 5 4 3
IROR, % 44.5 12.8 20.4 9.6 26.0
PI 1.89 0.92 1.13 0.89 1.19
PW at 15%, $1000 7,138 1,162 1,051 863 936
$18 million. She has confirmed the computations and is ready to do the ranking and make the
selection. Help her by doing the following:
(a) Use the IROR measure to rank and select projects.
(b) Use the PI measure to rank and select projects.
(c) Use the PW measure to rank and select projects.
(d) Compare the selected projects by the three methods and determine which one will maximize
the overall ROR value of the $18 million budget.
Solution
Refer to Table 12–5 for the ranking, cumulative investment for each project, and selection
based on the ranking and budget limit b $18 million.
(a) Ranking by overall IROR values indicates that projects 1 and 5 should be selected with
$13 million of the $18 million budget expended. The remaining $5 million is assumed to
be invested at the MARR of 15% per year.
(b) As an example, the PI for project 1 is calculated using Equation [12.7].
PI 1 ————————
4000(PA,15%,6)
|8000|
15,1388,000
1.89
Ranking and projects selected are the same as in the IROR-based analysis. Again, the remaining
$5 million is assumed to generate a return of MARR 15% per year.
(c) Ranking by PW value results in a different selection from that of IROR and PI ranking.
Projects 1 and 3, rather than 1 and 5, are selected for a total PW $8.189 million and an
investment of $16 million. The remaining $2 million is assumed to earn 15% per year.
(d) IROR and PI rankings result in projects 1 and 5 being selected. PW ranking results in the
selection of projects 1 and 3. With MARR 15%, the $18 million will earn at the
following ROR values. In $1000 units,
Projects 1 and 5 NCF, year 0: $13,000
NCF, years 1–3: $6,600
NCF, years 4–6: $4,000
0 13,000 6600(PA,i,3) 4000(PA,i,3)(PF,i,3)
TABLE 12–5 Ranking of Projects by Different Measures, Example 12.4 (Monetary units in $1000)
IROR, %
(1)
Ranking by IROR Ranking by PI Ranking by PW
Project
(2)
Cumulative
Investment, $
(3)
PI
(4)
Project
(5)
Cumulative
Investment, $
(6)
PW, $
(7)
Project
(8)
Cumulative
Investment, $
(9)
44.5 1 8,000 1.89 1 8,000 7,138 1 8,000
26.0 5 13,000 1.19 5 13,000 1,051 3 16,000
20.4 3 21,000 1.13 3 21,000 936 5 21,000
12.8 2 0.92 2 863 4
9.6 4 0.89 4 1,162 2

334 Chapter 12 Independent Projects with Budget Limitation
By IRR function, the rate of return is 39.1%. The overall return on the entire budget is
ROR [39.1(13,000) 15.0(5000)]18,000
32.4%
Projects 1 and 3 NCF, year 0: $16,000
NCF, years 1–5: $6,700
NCF, year 6: $4,000
0 16,000 6700(PA,i,5) 4000(PF,i,6)
By IRR function, the rate of return is 33.5%. The overall return on the entire budget is
ROR [33.5(16,000) 15.0(2000)]18,000
31.4%
In conclusion, the IROR and PI selections maximize the overall rate of return at 32.4%. The
PW selection maximizes the PW value at $8.189 (i.e., 7.138 1.051) million, as determined
from Table 12–5, column 7.
CHAPTER SUMMARY
Investment capital is always a scarce resource, so it must be rationed among competing projects
using economic and noneconomic criteria. Capital budgeting involves proposed projects, each
with an initial investment and net cash flows estimated over the life of the project. The fundamental
capital budgeting problem has specific characteristics (Figure 12–1).
• Selection is made from among independent projects.
• Each project must be accepted or rejected as a whole.
• Maximizing the present worth of the net cash flows is the objective.
• The total initial investment is limited to a specified maximum.
The present worth method is used for evaluation. To start the procedure, formulate all mutually
exclusive bundles that do not exceed the investment limit, including the do-nothing bundle.
There are a maximum of 2 m bundles for m projects. Calculate the PW at MARR for each bundle,
and select the bundle with the largest PW value. Reinvestment of net positive cash flows at the
MARR is assumed for all projects with lives shorter than that of the longest-lived project.
The capital budgeting problem may be formulated as a linear programming problem to select
projects directly in order to maximize the total PW. Excel’s Solver tool solves this problem by
spreadsheet.
Projects can be ranked and selected on bases other than PW values. Two measures are the
internal rate of return (IROR) and the profitability index (PI), also called the PW index. The
ordering of projects may differ between the various ranking bases since different measures are
optimized. When there are a large number of projects, the IROR basis is commonly applied in
industrial settings.
PROBLEMS
Understanding Capital Rationing
12.1 Define the following terms: bundle, contingent
project, dependent project.
12.2 State two assumptions made when doing capital rationing
using a PW analysis for unequal-life projects.
12.3 For independent projects identified as A, B, C, D,
E, F, and G, how many mutually exclusive bundles
can be formed?
12.4 For independent projects identified as W, X, Y, and
Z, develop all of the mutually exclusive bundles.
Projects X and Y perform the same function with
different processes; both should not be selected.
12.5 Five projects have been identified for possible
implementation by a company that makes dry ice
blasters—machines that propel tiny dry ice pellets
at supersonic speeds so they flash-freeze and
then lift grime, paint, rust, mold, asphalt, and

Problems 335
other contaminants off in-place machines and a
wide range of surfaces. The total present worth
of the initial investment for each project is
shown. Determine which bundles are possible,
provided the budget limitation is ( a ) $34,000 and
( b ) $45,000.
Project
PW of Investment
at 15%, $
L 28,000
M 11,000
N 43,000
O 38,000
P 6,000
12.6 Four independent projects (1, 2, 3, and 4) are proposed
for investment by Perfect Manufacturing,
Inc. List all the acceptable mutually exclusive
bundles based on the following selection restrictions
developed by the department of engineering
production:
Project 2 can be selected only if project 3 is
selected.
Projects 1 and 4 should not both be selected;
they are essentially duplicates.
12.7 Develop all acceptable mutually exclusive bundles
for the four independent projects described
below if the investment limit is $400 and the following
project selection restriction applies: Project
1 can be selected only if both projects 3 and 4
are selected.
Project Investment, $
1 100
2 150
3 75
4 235
Selection from Independent Projects
12.8 Listed below are bundles, each comprised of three
independent proposals for which the PW has been
estimated. Select the best bundle if the capital budget
limit is $45,000 and the MARR is the cost of
capital, which is 9% per year.
Proposal
Bundle
Initial Investment,
For Bundle, $
PW at 9%,
$
1 18,000 1,400
2 26,000 8,500
3 34,000 7,100
4 41,000 10,500
12.9 The general manager for Woodslome Appliance
Company Plant #A14 in Mexico City has four
independent projects that she can fund this year to
improve surface durability on stainless steel
products. The project costs and 18% per year PW
values are as shown. What projects should be accepted
if the budget limit is ( a ) no limit and
( b ) $55,000?
Project
Initial
Investment, $
PW at 18%,
$
1 15,000 400
2 25,000 8500
3 20,000 500
4 40,000 9600
12.10 The capital fund for research project investment at
SummaCorp is limited to $100,000 for next year.
Select any or all of the following proposals if a
MARR of 15% per year is established by the board
of directors.
Project
Initial
Investment, $
Annual Net Cash
Flow, $Year
Life,
Years
Salvage
Value, $
I 25,000 6,000 4 4,000
II 30,000 9,000 4 1,000
III 50,000 15,000 4 20,000
12.11 An electrical engineer at GE is assigned the responsibility
to determine how to invest up to $100,000
in none, some, or all of the following independent
proposals. Use ( a ) hand and ( b ) spreadsheet-based
PW analysis and a 15% per year return requirement
to help this engineer make the best decision from a
purely economic perspective.
Project
Initial
Investment, $
Annual Net Cash
Flow, $Year
Life,
Years
Salvage
Value, $
A 25,000 6,000 4 4,000
B 20,000 9,000 4 0
C 50,000 15,000 4 20,000
12.12 Dwayne has four independent vendor proposals to
contract the nationwide oil recycling services for
Ford Corporation manufacturing plants. All combinations
are acceptable, except that vendors B
and C cannot both be chosen. Revenue sharing of
recycled oil sales with Ford is a part of the requirement.
Develop all possible mutually exclusive
bundles under the additional following restrictions
and select the best projects. The corporate MARR
is 10% per year.
(a) A maximum of $4 million can be spent.
( b) A larger budget of $5.5 million is allowed,
but no more than two vendors can be
selected.
(c) There is no limit on spending.

336 Chapter 12 Independent Projects with Budget Limitation
Vendor
Initial
Investment, $
Life,
Years
Annual Net
Revenue, $ per Year
A 1.5 million 8 360,000
B 3.0 million 10 600,000
C 1.8 million 5 620,000
D 2.0 million 4 630,000
12.13 Use the PW method at 8% per year to select up to
three projects from the four available ones if no
more than $20,000 can be invested. Estimated
lives and annual net cash flows vary.
Project
Initial
Investment, $
Net Cash Flow, $ per Year
1 2 3 4 5 6
W 5,000 1,000 1,700 1,800 2,500 2,000
X 8,000 900 950 1,000 1,050 10,500
Y 8,000 4,000 3,000 1,000 500 500 2,000
Z 10,000 0 0 0 17,000
12.14 Charlie’s Garage has $70,000 to spend on new
equipment that may increase revenue for his car
repair shop. Use the PW method to determine
which of these independent investments are financially
acceptable at 6% per year, compounded
monthly. All are expected to last 3 years.
Feature
Installed
Cost, $
Estimated Added
Revenue, $ per Month
Diagnostics system 45,000 2200
Exhaust analyzer 30,000 2000
Hybrid engine tester 22,000 1500
Project
Initial
Investment, $
Life, Years
PW at 12%
per Year, $
S 15,000 6 8,540
A 25,000 8 12,325
M 10,000 6 3,000
E 25,000 4 10
H 40,000 12 15,350
12.17 The independent project estimates below have
been developed by the engineering and finance
managers. The corporate MARR is 8% per year,
and the capital investment limit is $4 million. Select
the economically best projects using the PW
method and ( a ) hand solution and ( b ) spreadsheet
solution.
Project
Project
Cost, $ M
Life, Years
NCF,
$ per Year
1 1.5 8 360,000
2 3.0 10 600,000
3 1.8 5 520,000
4 2.0 4 820,000
12.18 Use the PW method to evaluate four independent
projects. Select as many as three of the four projects.
The MARR is 12% per year, and up to
$16,000 in capital investment funds are available.
Project
1 2 3 4
Investment, $ 5000 8,000 9,000 10,000
Life, years 5 5 3 4
12.15 (a) Determine which of the following independent
projects should be selected for investment
if $315,000 is available and the MARR
is 10% per year. Use the PW method to
evaluate mutually exclusive bundles to
make the selection. (Solve by hand or
spreadsheet as instructed.)
Project
Initial
Investment, $
NCF,
$ per Year Life, Years
A 100,000 50,000 8
B 125,000 24,000 8
C 120,000 75,000 8
D 220,000 39,000 8
E 200,000 82,000 8
(b)
If the five projects are mutually exclusive
alternatives, perform the present worth analysis
and select the best alternative.
12.16 Use the following analysis of five independent
projects to select the best, if the capital limitation
is ( a ) $30,000, ( b ) $52,000, and ( c ) unlimited.
Year
NCF Estimates, $ per Year
1 1000 500 5000 0
2 1700 500 5000 0
3 2400 500 2000 0
4 3000 500 17,000
5 3800 10,500
12.19 Work Problem 12.18 using a spreadsheet.
12.20 A capital rationing problem is defined for you as
follows: Three projects are to be evaluated at a
MARR of 12.5% per year. No more than $3.0
million can be invested.
(a) Use a spreadsheet to select from the independent
projects.
(b) If the life of project 3 can be increased from
5 to 10 years for the same $1 million investment,
use Goal Seek to determine the NCF in
year 1 for project 3 alone to have the same
PW as the best bundle in part ( a ). All other
estimates remain the same. With these new
NCF and life estimates, what are the best
projects for investment?

Problems 337
Project
Investment,
$ M
Life,
Years Year 1
Estimated NCF, $ per Year
Gradient
after Year 1
1 0.9 6 250,000 5000
2 2.1 10 485,000 5000
3 1.0 5 200,000 20%
Linear Programming and Capital Budgeting
12.21 Formulate the linear programming model, develop
a spreadsheet, and solve the capital rationing problem
in Example 12.1 ( a ) as presented and ( b ) using
an investment limit of $13 million.
12.22 Use linear programming and a spreadsheet-based
solution technique to select from the independent
unequal-life projects in Problem 12.17.
12.23 Solve the capital budgeting problem in Problem
12.20( a ), using the linear programming model and
a spreadsheet.
12.24 Johnson and Johnson is expanding its first-aid
products line for individuals allergic to latex. The
research team comprised of doctors, engineers,
and chemists has proposed the four projects estimated
in Problem 12.18. Develop the linear
programming model and use a spreadsheet to select
the projects to be funded. The MARR is 12%
per year, and the budget limit is $16 million. All
monetary values are in $1000 units.
12.25 Using the estimates in Problem 12.18 and repeated
spreadsheet solution of the capital budgeting problem
for budget limits ranging from b $5000 to b
$25,000, develop a spreadsheet chart that plots b
versus the value of Z .
Other Ranking Measures
12.26 A newly proposed project has a first cost of
$325,000 and estimated annual income of $60,000
per year for 8 years.
( a) Determine the IROR, PI, and PW values if
the MARR is 15% per year.
( b) Is the project economically justified?
12.27 An engineer at Delphi Systems is considering the
projects below, all of which can be considered to
last indefinitely. The company’s MARR is 13%
per year.
( a) Determine which projects should be selected
on the basis of IROR if the budget limitation
is $39,000.
( b) What is the overall rate of return if the money
not invested in projects is assumed to earn a
rate of return equal to the MARR?
Project First Cost, $
Net Income,
$ per Year
Rate of
Return, %
A 20,000 4,000 20.0
B 10,000 1,900 19.0
C 15,000 2,600 17.3
D 70,000 10,000 14.3
E 50,000 6, 000 12.0
12.28 The five independent projects shown below are under
consideration for implementation by KNF Neuberger,
Inc. The company’s MARR is 15% per year.
(a) Determine which projects should be undertaken
on the basis of IROR if the budget
limitation is $97,000. (Solve by hand or
spreadsheet as instructed.)
(b) Determine the overall rate of return if the
money not invested in projects is assumed to
earn a rate of return equal to the MARR.
Project First Cost, $
Annual
Income, $ per Year
Project
Life, Years
L 30,000 9,000 10
A 15,000 4,900 10
N 45,000 11,100 10
D 70,000 9,000 10
T 40,000 10,000 10
12.29 An estimated 6 billion gallons of clean drinking
water disappear each day across the United States
due to aging, leaky pipes and water mains before
it gets to the consumer or industrial user. The American
Society of Civil Engineers (ASCE) has teamed
with municipalities, counties, and several excavation
companies to develop robots that can travel through
mains, detect leaks, and repair many of them immediately.
Four proposals have been received for funding.
There is a $100 million limit on capital funding,
and the MARR is established at 12% per year.
(a) Use the IROR method to rank and determine
which of the four independent projects should
be funded. Solve by spreadsheet, unless instructed
to use hand solution.
Project
(b)
Determine the rate of return for the combination
of projects selected in part ( a ). Is it economically
justified?
(c) Determine the overall rate of return for the
$100 million with the projects selected in
part ( a ). Assume that excess funds are invested
at the MARR.
First
Cost, $ M
Estimated Annual
Savings, $M per Year
Project
Life, Years
W 12 5.0 3
X 25 7.3 4
Y 45 12.1 6
Z 60 9.0 8

338 Chapter 12 Independent Projects with Budget Limitation
12.30 Determine the profitability index at 10% per year
interest for a project that has a first cost of
$200,000 in year 0 and $200,000 in year 2, an annual
operating cost of $80,000 per year, revenue
of $170,000 per year, and a salvage value of
$60,000 after its 5-year life.
12.31 The budget limit is $120,000, and the interest rate
is 10% per year. All projects have a 10-year life.
Use ( a ) the PI method and ( b ) the IROR method to
rank and select from the independent projects.
( c ) Are different projects selected using the two
methods?
Project First Cost, $
Net Income,
$ per Year IROR, %
A 18,000 4,000 18.0
B 15,000 2,800 13.3
C 35,000 12,600 34.1
D 60,000 13,000 17.3
E 50,000 8,000 9.6
12.32 The six independent projects shown below are
under consideration by Peyton Packing under
budget-constrained conditions. The company always
has more projects to engage in than it has
capital to fund projects. Therefore, it uses a relatively
high MARR of 25% per year. Since all projects
are considered long-term ventures, the company
uses an infinite period for their life. Determine
which projects the company should fund and the
total investment for a capital budget of $700,000 if
the capital budgeting method used is ( a ) the IROR
method, ( b ) the PI method, and ( c ) the PW method.
Project First Cost, $
Estimated Annual
Income, $ per Year
F 200,000 54,000
G 120,000 21,000
H 250,000 115,000
I 370,000 205,000
J 50,000 26,000
K 9,000 2,100
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
12.33 A project that has a condition associated with its
acceptance or rejection is known as:
(a) A mutually exclusive alternative
(b) A contingent project
(c) A dependent project
(d) Both ( b ) and ( c )
(a)
(b)
(c)
(d)
One of the bundles is the do-nothing project.
A bundle may consist of only one project.
The capital limit may be exceeded as long as
it is exceeded by less than 3%.
A bundle may include contingent and dependent
projects.
12.34 An assumption inherent in capital budgeting is that
all positive net cash flows are reinvested at the
MARR from the time they are realized until:
(a) The end of the shortest-lived project
(b)
(c)
The end of the longest-lived project
The end of the respective project that generated
the cash flow
(d) The average of the lives of the projects included
in the bundle
12.35 All of the following are correct when a capital
budgeting problem is solved using the 0-1 integer
linear programming model except:
(a) Partial investment in a project is acceptable.
(b) The objective is to maximize the present
worth of the investments.
(c) Budget constraints may be present for the
first year only or for several years.
(d) Contingent and dependent project restrictions
may be considered.
12.36 All of the following are true when formulating
mutually exclusive bundles of independent projects
except:
12.37 When there are 5 projects involved in a capital
budgeting study, the maximum number of bundles
that can be formulated is:
( a ) 6
( b ) 10
( c ) 31
( d ) 32
12.38 For a capital limit of $25,000, the selected independent
projects are:
( a ) P only
( b ) Q only
( c ) R only
( d ) P and R
Project
Initial
Investment, $
Life, Years
PW at
12%, $
P 15,000 6 8,540
Q 25,000 8 12,325
R 10,000 6 3,000
S 25,000 4 10
T 40,000 12 15,350

Additional Problems and FE Exam Review Questions 339
12.39 The independent projects shown below are under
consideration for possible implementation by
Renishaw Inc. of Hoffman Estates, Rhode Island.
If the company’s MARR is 14% per year and it
uses the IROR method of capital budgeting, the
projects it should select under a budget limitation
of $105,000 are:
( a) A, B, and C
( b) A, B, and D
( c) B, C, and D
( d) A, C, and D
12.40 For a project that requires an initial investment
of $26,000 and yields $10,000 per year for
4 years, the PI at an interest rate of 10% per year
is closest to:
( a ) 1.03
( b ) 1.22
( c ) 1.38
( d ) 1.56
Project First Cost, $
Annual
Income, $ per Year
Rate of
Return, %
A 20,000 4,000 20.0
B 10,000 1,900 19.0
C 15,000 2,600 17.3
D 70,000 10,000 14.3
E 50,000 6,000 12.0

CHAPTER 13
Breakeven and
Payback
Analysis
LEARNING OUTCOMES
Purpose: Determine the breakeven for one or two alternatives and calculate the payback period with and without a
return required.
SECTION TOPIC LEARNING OUTCOME
13.1 Breakeven point • Determine the breakeven point for one
parameter.
13.2 Two alternatives • Calculate the breakeven point of a parameter
and use it to select between two alternatives.
13.3 Payback period • Determine the payback period of a project at
i 0% and i 0%. Illustrate the cautions when
using payback analysis.
13.4 Spreadsheet • Answer breakeven and payback questions that
are best resolved by spreadsheet and Goal Seek
tools.

B
reakeven analysis is performed to determine the value of a variable or parameter
of a project or alternative that makes two elements equal, for example, the
sales volume that will equate revenues and costs. A breakeven study is performed
for two alternatives to determine when either alternative is equally acceptable. Breakeven
analysis is commonly applied in make-or-buy decisions when a decision is needed about the
source for manufactured components, services, etc.
Payback analysis determines the required minimum life of an asset, process, or system to
recover the initial investment. There are two types of payback: return ( i 0%) and no return
( i 0%). Payback analysis should not be considered the final decision maker; it is used as
a screening tool or to provide supplemental information for a PW, AW, or other analysis.
These aspects are discussed in depth in this chapter.
Breakeven and payback studies use estimates that are considered to be certain; that is, if
the estimated values are expected to vary enough to possibly change the outcome, another
study is necessary using different estimates. If the variable of interest is allowed to vary, the
approaches of sensitivity analysis (Chapter 18) should be used. Additionally, if probability
and risk assessment are considered, the tools of simulation (Chapter 19) can be used to
supplement the static nature of a breakeven or payback study.
13.1 Breakeven Analysis for a Single Project
When one of the engineering economy symbols— P , F , A , i , or n —is not known or not estimated,
a breakeven quantity can be determined by setting an equivalence relation for PW or AW equal
to zero. This form of breakeven analysis has been used many times so far. For example, we have
solved for the rate of return i *, found the replacement value for a defender, and determined the P ,
F , A , or salvage value S at which a series of cash flow estimates return a specific MARR. Methods
used to determine the quantity include
Direct solution by hand if only one factor is present (say, P A ) or only single amounts are
estimated (for example, P and F )
Trial and error by hand or calculator when multiple factors are present
Spreadsheet when cash flow and other estimates are entered into cells and used in resident
functions (PV, FV, RATE, IRR, NPV, PMT, and NPER) or tools (Goal Seek and Solver).
We now concentrate on the determination of the breakeven quantity Q BE for one parameter
or decision variable. For example, the variable may be a design element to minimize cost or the
production level needed to realize revenues that exceed costs by 10%.
Breakeven analysis finds the value of a parameter that makes two elements equal . The breakeven
point Q BE is determined from mathematical relations, e.g., product revenue and costs or
materials supply and demand or other parameters that involve the parameter Q . Breakeven
analysis is fundamental to evaluations such as make-buy decisions.
Breakeven
The unit of the parameter Q may vary widely: units per year, cost per kilogram, hours per month,
percentage of full plant capacity, etc.
Figure 13–1 a presents different shapes of a revenue relation identified as R . A linear revenue
relation is commonly assumed, but a nonlinear relation is often more realistic. It can model an
increasing per unit revenue with larger volumes (curve 1 in Figure 13–1 a ) or a decreasing per
unit revenue that usually prevails at higher quantities (curve 2).
Costs, which may be linear or nonlinear, usually include two components—fixed and
variable—as indicated in Figure 13–1 b .
Fixed costs (FC). These include costs such as buildings, insurance, fixed overhead, some
minimum level of labor, equipment capital recovery, and information systems.
Variable costs (VC). These include costs such as direct labor, materials, indirect costs, contractors,
marketing, advertisement, and warranty.
The fixed-cost component is essentially constant for all values of the variable, so it does not vary
for a large range of operating parameters, such as production level or workforce size. Even if no
units are produced, fixed costs are incurred at some threshold level. Of course, this situation

342 Chapter 13 Breakeven and Payback Analysis
Figure 13–1
Linear and nonlinear
revenue and cost relations.
R, revenue per year
(2)
(1)
Linear
Nonlinear
Q, units per year
(a) Revenue relations—(1) increasing and
(2) decreasing revenue per unit
TC FC VC
TC FC VC
Cost per year
Total
cost, TC
Variable, VC
Cost per year
TC
VC
Fixed, FC
Q, units per year
(b) Linear cost relations
FC
Q, units per year
(c) Nonlinear cost relations
cannot last long before the plant must shut down to reduce fixed costs. Fixed costs are reduced
through improved equipment, information systems, and workforce utilization; and less costly
fringe benefit packages, subcontracting specific functions, and so on.
Variable costs change with production level, workforce size, and other parameters. It is usually
possible to decrease variable costs through better product design, manufacturing efficiency,
improved quality and safety, and higher sales volume.
When FC and VC are added, they form the total cost relation TC . Figure 13–1 b illustrates
the TC relation for linear fixed and variable costs. Figure 13–1 c shows a general TC curve for a
nonlinear VC in which unit variable costs decrease as the quantity level rises.
At a specific but unknown value Q of the decision variable, the revenue R and total cost TC
relations will intersect to identify the breakeven point Q BE (Figure 13–2). If Q Q BE , there is a
predictable profit ; but if Q Q BE , there is a loss. For linear models of R and VC, the greater the
quantity, the larger the profit. Profit is calculated as
Profit revenue total cost
R TC
R (FC VC) [13.1]
A relation for the breakeven point may be derived when revenue and total cost are linear functions
of quantity Q by setting the relations for R and TC equal to each other, indicating a profit of zero.
R TC
rQ FC vQ
where
r revenue per unit
v variable cost per unit
Solve for the breakeven quantity Q Q BE for linear R and TC functions.
Q BE FC ———
r v
[13.2]

13.1 Breakeven Analysis for a Single Project 343
$
R
Breakeven
TC
Breakeven with
lowered VC
TC with
lowered VC
Profit
maximized
Loss
Breakeven
point moves
Profit
Q BE
Q, units per year
Figure 13–2
Effect on the breakeven point when the variable cost per unit is reduced.
$
TC
Profit
maximized
R
Profit range
Loss
Loss
Q BE
Q P Q BE
Q, units per year
Figure 13–3
Breakeven points and maximum profit point for a nonlinear analysis.
The breakeven graph is an important management tool because it is easy to understand and may be used
in decision making in a variety of ways. For example, if the variable cost per unit is reduced, then the
TC line has a smaller slope (Figure13–2) and the breakeven point will decrease. This is an advantage
because the smaller the value of Q BE , the greater the profit for a given amount of revenue. A similar
analysis is possible for fixed VC and increased levels of production, as shown in the next example.
If nonlinear R or TC models are used, there may be more than one breakeven point. Figure 13–3
presents this situation for two breakeven points. The maximum profit occurs at Q P between the
two breakeven points where the distance between the R and TC relations is greatest.

344 Chapter 13 Breakeven and Payback Analysis
EXAMPLE 13.1
Of course, no static R and TC relations—linear or nonlinear—are able to estimate exactly the
revenue and cost amounts over an extended period of time. But the breakeven point is an excellent
target for planning purposes.
Indira Industries is a major producer of diverter dampers used in the gas turbine power industry
to divert gas exhausts from the turbine to a side stack, thus reducing the noise to acceptable
levels for human environments. Normal production level is 60 diverter systems per month, but
due to significantly improved economic conditions in Asia, production is at 72 per month. The
following information is available.
Fixed costs
FC $2.4 million per month
Variable cost per unit v $35,000
Revenue per unit r $75,000
(a) How does the increased production level of 72 units per month compare with the current
breakeven point?
(b) What is the current profit level per month for the facility?
(c) What is the difference between the revenue and variable cost per damper that is necessary
to break even at a significantly reduced monthly production level of 45 units, if fixed costs
remain constant?
Solution
(a) Use Equation [13.2] to determine the breakeven number of units. All dollar amounts are
in $1000 units.
Q BE ———
r FC v
———— 2400
60 units per month
75 35
Figure 13–4 is a plot of R and TC lines. The breakeven value is 60 damper units. The
increased production level of 72 units is above the breakeven value.
7000
Revenue line
6000
$1000
5000
4500
4000
3000
2400
2000
Total cost line
Profit
1000
Q BE = 60
0 15 30 45 60 75 90
Q, units per month
Figure 13–4
Breakeven graph, Example 13.1.

13.2 Breakeven Analysis Between Two Alternatives 345
(b) To estimate profit (in $1000 units) at Q 72 units per month, use Equation [13.1].
Profit R TC rQ (FC vQ)
(r v)Q FC
(75 35)72 2400 [13.3]
$480
There is a profit of $480,000 per month currently.
(c) To determine the required difference r v, use Equation [13.3] with profit 0, Q 45,
and FC $2.4 million. In $1000 units,
0 (r v) (45) 2400
r v ——— 2400 $53.33 per unit
45
The spread between r and v must be $53,330. If v stays at $35,000, the revenue per damper
must increase from $75,000 to $88,330 (i.e., 35,000 53,330) just to break even at a production
level of Q 45 per month.
In some circumstances, breakeven analysis performed on a per unit basis is more meaningful.
The value of Q BE is still calculated using Equation [13.2], but the TC relation is divided by Q to
obtain an expression for cost per unit, also termed average cost per unit C u .
C u TC ——
Q
FC vQ ————
Q
FC ——
Q v [13.4]
At the breakeven quantity Q Q BE , the revenue per unit is exactly equal to the cost per unit. If
graphed, the FC per unit term in Equation [13.4] takes on the shape of a hyperbola.
The breakeven point for a project with one unknown variable can always be determined by
equating revenue and total cost. This is the same as setting profit equal to zero in Equation [13.1].
It may be necessary to perform some dimensional analysis initially to obtain the correct revenue
and total cost relations in order to use the same dimension for both relations, for example, $ per
unit, miles per month, or units per year.
13.2 Breakeven Analysis Between Two Alternatives
Now we consider breakeven analysis between two mutually exclusive alternatives.
Breakeven analysis determines the value of a common variable or parameter between two
alternatives. Equating the two PW or AW relations determines the breakeven point. Selection
of the alternative is different depending upon two facts: slope of the variable cost curve and
the parameter value relative to the breakeven point .
Breakeven
The parameter can be the interest rate i , first cost P , annual operating cost (AOC), or any parameter.
We have already performed breakeven analysis between alternatives on several parameters.
For example, the incremental ROR value ( i *) is the breakeven rate between alternatives. If the
MARR is lower than i *, the extra investment of the larger-investment alternative is justified. In
Section 11.6, the replacement value (RV) of a defender was determined. If the market value is
larger than RV, the decision should favor the challenger.
Often breakeven analysis involves revenue or cost variables common to both alternatives,
such as price per unit, operating cost, cost of materials, or labor cost. Figure 13–5 illustrates
the breakeven concept for two alternatives with linear cost relations. The fixed cost of alternative
2 is greater than that of alternative 1. However, alternative 2 has a smaller variable cost,
as indicated by its lower slope. The intersection of the total cost lines locates the breakeven
point, and the variable cost establishes the slope. Thus, if the number of units of the common
variable is greater than the breakeven amount, alternative 2 is selected, since the total cost
will be lower. Conversely, an anticipated level of operation below the breakeven point favors
alternative 1.

346 Chapter 13 Breakeven and Payback Analysis
Figure 13–5
Breakeven between two
alternatives with linear
cost relations.
Total cost
Alt. 2 FC
Alt. 1 TC
Alt. 2 TC
Alt. 1 FC
Breakeven
point
Common variable, units
Instead of plotting the total costs of each alternative and estimating the breakeven point graphically,
it may be easier to calculate the breakeven point numerically using engineering economy
expressions for the PW or AW at the MARR. The AW is preferred when the variable units are
expressed on a yearly basis, and AW calculations are simpler for alternatives with unequal lives.
The following steps determine the breakeven point of the common variable and the slope of a
linear total cost relation.
1. Define the common variable and its dimensional units.
2. Develop the PW or AW relation for each alternative as a function of the common variable.
3. Equate the two relations and solve for the breakeven value of the variable.
Selection between alternatives is based on this guideline:
If the anticipated level of the common variable is below the breakeven value, select the alternative
with the higher variable cost (larger slope).
If the level is above the breakeven point, select the alternative with the lower variable cost.
(Refer to Figure 13–5.)
EXAMPLE 13.2
A small aerospace company is evaluating two alternatives: the purchase of an automatic feed
machine and a manual feed machine for a finishing process. The auto feed machine has an
initial cost of $23,000, an estimated salvage value of $4000, and a predicted life of 10 years.
One person will operate the machine at a rate of $12 per hour. The expected output is 8 tons per
hour. Annual maintenance and operating cost is expected to be $3500.
The alternative manual feed machine has a first cost of $8000, no expected salvage value, a
5-year life, and an output of 6 tons per hour. However, three workers will be required at $8 per
hour each. The machine will have an annual maintenance and operation cost of $1500. All
projects are expected to generate a return of 10% per year. How many tons per year must be
finished to justify the higher purchase cost of the auto feed machine?
Solution
Use the steps above to calculate the breakeven point between the two alternatives.
1. Let x represent the number of tons per year.
2. For the auto feed machine, the annual variable cost is
Annual VC ——
$12
hour
1.5x
1 hour ———
8 tons
x tons ———
year
The VC is developed in dollars per year. The AW expression for the auto feed
machine is
AW auto 23,000(AP,10%,10) 4000(AF,10%,10) 3500 1.5x
$–6992 1.5x

13.2 Breakeven Analysis Between Two Alternatives 347
Similarly, the annual variable cost and AW for the manual feed machine are
Annual VC ——
$8
hour
4x
(3 operators)
1 hour ———
6 tons
x tons ———
year
AW manual 8000(AP,10%,5) 1500 4x
$3610 4x
3. Equate the two cost relations and solve for x.
AW auto AW manual
6992 1.5x 3610 – 4x
x 1353 tons per year
If the output is expected to exceed 1353 tons per year, purchase the auto feed machine, since
its VC slope of 1.5 is smaller than the manual feed VC slope of 4.
The breakeven analysis approach is commonly used for make-or-buy decisions . This means
the company contracts to buy the product or service from the outside , or makes it within the company.
The alternative to buy usually has no fixed cost and a larger variable cost than the option to
make. Where the two cost relations cross is the make-buy decision quantity. Amounts above this
indicate that the item should be made, not purchased outside.
EXAMPLE 13.3
Guardian is a national manufacturing company of home health care appliances. It is faced with
a make-or-buy decision. A newly engineered lift can be installed in a car trunk to raise and
lower a wheelchair. The steel arm of the lift can be purchased internationally for $3.50 per unit
or made in-house. If manufactured on site, two machines will be required. Machine A is estimated
to cost $18,000, have a life of 6 years, and have a $2000 salvage value; machine B will
cost $12,000, have a life of 4 years, and have a $500 salvage value (carry-away cost). Machine
A will require an overhaul after 3 years costing $3000. The annual operating cost for
machine A is expected to be $6000 per year and for machine B is $5000 per year. A total of four
operators will be required for the two machines at a rate of $12.50 per hour per operator. In a
normal 8-hour period, the operators and two machines can produce parts sufficient to manufacture
1000 units. Use a MARR of 15% per year to determine the following.
(a) Number of units to manufacture each year to justify the in-house (make) option.
(b) The maximum capital expense justifiable to purchase machine A, assuming all other estimates
for machines A and B are as stated. The company expects to produce 10,000 units
per year.
Solution
(a) Use steps 1 to 3 stated previously to determine the breakeven point.
1. Define x as the number of lifts produced per year.
2. There are variable costs for the operators and fixed costs for the two machines for the
make option.
Annual VC (cost per unit)(units per year)
—————
4 operators $12.50
———(8 hours)x
1000 units hour
0.4x
The annual fixed costs for machines A and B are the AW amounts.
AW A 18,000(AP,15%,6) 2000(AF,15%,6)
6000 3000(PF,15%,3)(AP,15%,6)
AW B 12,000(AP,15%,4) 500(AF,15%,4) 5000
Total cost is the sum of AW A , AW B , and VC.

348 Chapter 13 Breakeven and Payback Analysis
3. Equating the annual costs of the buy option (3.50x) and the make option yields
3.50x AW A AW B VC
18,000(AP,15%,6) 2000(AF,15%,6) 6000
3000(PF,15%,3)(AP,15%,6) 12,000(AP,15%,4)
500(AF,15%,4) 5000 0.4x [13.5]
3.10x 20,352
x 6565 units per year
A minimum of 6565 lifts must be produced each year to justify the make option, which
has the lower variable cost of 0.4x.
(b) Substitute 10,000 for x and P A for the to-be-determined first cost of machine A (currently
$18,000) in Equation [13.5]. Solution yields P A $58,295. This is approximately three
times the estimated first cost of $18,000, because the production of 10,000 per year is
considerably larger than the breakeven amount of 6565.
Even though the preceding examples treat only two alternatives, the same type of analysis can
be performed for three or more alternatives. To do so, compare the alternatives in pairs to find
their respective breakeven points. The results are the ranges through which each alternative is
more economical. For example, in Figure 13–6, if the output is less than 40 units per hour, alternative
1 should be selected. Between 40 and 60, alternative 2 is more economical; and above 60,
alternative 3 is favored.
If the variable cost relations are nonlinear, analysis is more complicated. If the costs increase
or decrease uniformly, mathematical expressions that allow direct determination of the breakeven
point can be developed.
13.3 Payback Analysis
Payback analysis is another use of the present worth technique. It is used to determine the amount
of time, usually expressed in years, required to recover the first cost of an asset or project. Payback
is allied with breakeven analysis; this is illustrated later in the section. The payback period ,
also called payback or payout period , has the following definition and types.
Alternative 1
Alternative 2
Alternative 3
Total cost, $/year
Breakeven
points
Figure 13–6
Breakeven points for three alternatives.
40
60
Output, units/hour

13.3 Payback Analysis 349
The payback period n p is an estimated time for the revenues, savings, and any other monetary
benefits to completely recover the initial investment plus a stated rate of return i.
There are two types of payback analysis as determined by the required return.
No return; i0%: Also called simple payback, this is the recovery of only the initial investment.
Discounted payback; i 0%: The time value of money is considered in that some return, for
example, 10% per year, must be realized in addition to recovering the initial investment.
Payback period
An example application of payback may be a corporate senior manager who insists that every
proposal return the initial cost and some stated return within 3 years. Using payback as an initial
screening tool, no proposal with n p 3 years can become a viable alternative. The payback period
should be determined using a required i 0%. Unfortunately in practice, no-return payback
is used too often to make economic decisions. After the formulas are presented, a couple of cautions
about payback usage are provided.
The equations used to determine n p differ for each type of analysis. For both types, the terminology
is P for the initial investment in the asset, project, contract, etc., and NCF for the estimated
annual net cash flow. Using Equation [1.5], annual NCF is
NCF cash inflows cash outflows
To calculate the payback period for i 0% or i 0%, determine the pattern of the NCF series.
Note that n p is usually not an integer. For t 1, 2, . . . , n p ,
No return, i 0%; NCF t varies annually:
No return, i 0%; annual uniform NCF:
tn p
0 P NCF t [13.6]
t1
n p ——— P
[13.7]
NCF
Discounted, i 0%; NCF t varies annually:
tn p
0 P NCF t (PF, i, t) [13.8]
t1
Discounted, i 0%; annual uniform NCF: 0 P NCF(PA, i, n p ) [13.9]
After n p years, the cash flows will recover the investment in year 0 plus the required return of
i %. If the alternative is used more than n p years, with the same or similar cash flows, a larger
return results. If the estimated life is less than n p years, there is not enough time to recover the
investment and i % return. It is important to understand that payback analysis neglects all cash
fl ows after the payback period of n p years. Consequently, it is preferable to use payback as an
initial screening method or supplemental tool rather than as the primary means to select an
alternative. The reasons for this caution are that
• No-return payback neglects the time value of money, since no return on an investment is required.
• Either type of payback disregards all cash flows occurring after the payback period . These
cash flows may increase the return on the initial investment.
Payback analysis utilizes a significantly different approach to alternative evaluation than the
primary methods of PW, AW, ROR, and BC. It is possible for payback analysis to select a different
alternative than these techniques. However, the information obtained from discounted payback
analysis performed at an appropriate i 0% can be very useful in that a sense of the risk
involved in undertaking an alternative is provided. For example, if a company plans to utilize a
machine for only 3 years and payback is 6 years, indication is that the equipment should not be
obtained. Even here, the 6-year payback is considered supplemental information and does not
replace a complete economic analysis.
EXAMPLE 13.4
The board of directors of Halliburton International has just approved an $18 million worldwide
engineering construction design contract. The services are expected to generate new
annual net cash flows of $3 million. The contract has a potentially lucrative repayment clause

350 Chapter 13 Breakeven and Payback Analysis
to Halliburton of $3 million at any time that the contract is canceled by either party during the
10 years of the contract period. (a) If i 15%, compute the payback period. (b) Determine
the no-return payback period and compare it with the answer for i 15%. This is an initial
check to determine if the board made a good economic decision. Show both hand and spreadsheet
solutions.
Solution by Hand
(a) The net cash flow each year is $3 million. The single $3 million payment (call it CV for
cancellation value) could be received at any time within the 10-year contract period.
Equation [13.9] is altered to include CV.
In $1,000,000 units,
0 P NCF(PA,i,n) CV(PF,i,n)
0 18 3(PA,15%,n) 3(PF,15%,n)
The 15% payback period is n p 15.3 years, found by trail and error. During the period of
10 years, the contract will not deliver the required return.
(b) If Halliburton requires absolutely no return on its $18 million investment, Equation [13.6]
results in n p 5 years, as follows (in $ million).
0 18 5(3) 3
There is a very significant difference in n p for 15% and 0%. At 15% this contract would
have to be in force for 15.3 years, while the no-return payback period requires only 5 years.
A longer time is always required for i 0% for the obvious reason that the time value of
money is considered.
Solution by Spreadsheet
Enter the function NPER(15%,3,18,3) to display 15.3 years. Change the rate from 15% to
0% to display the no-return payback period of 5 years.
If two or more alternatives are evaluated using payback periods to indicate that one may
be better than the other(s), the second shortcoming of payback analysis (neglect of cash
flows after n p ) may lead to an economically incorrect decision. When cash flows that occur
after n p are neglected, it is possible to favor short-lived assets even when longer-lived assets
produce a higher return. In these cases, PW (or AW) analysis should always be the primary
selection method. Comparison of short- and long-lived assets in Example 13.5 illustrates
this situation.
EXAMPLE 13.5
Two equivalent pieces of quality inspection equipment are being considered for purchase by
Square D Electric. Machine 2 is expected to be versatile and technologically advanced enough
to provide net income longer than machine 1.
Machine 1 Machine 2
First cost, $ 12,000 8,000
Annual NCF, $ 3,000 1,000 (years 1–5),
3,000 (years 6–14)
Maximum life, years 7 14
The quality manager used a return of 15% per year and software that incorporates Equations
[13.8] and [13.9] to recommend machine 1 because it has a shorter payback period of
6.57 years at i 15%. The computations are summarized here.

13.3 Payback Analysis 351
$3000 per year
0 1 2 3 4 5 6 7
Cash flow neglected
by payback analysis
Machine 1 n p = 6.57
$12,000
$1000 per year
$3000 per year
Cash flows neglected
by payback analysis
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Machine 2 n p = 9.52
$8000
Figure 13–7
Illustration of payback periods and neglected net cash flows, Example 13.5.
Machine 1: n p 6.57 years, which is less than the 7-year life.
Equation used: 0 12,000 3000(PA,15%,n p )
Machine 2: n p 9.52 years, which is less than the 14-year life.
Equation used:
Recommendation: Select machine 1.
0 8000 1000(PA,15%,5)
3000(PA,15%,n p 5)(PF,15%,5)
Now, use a 15% PW analysis to compare the machines and comment on any difference in the
recommendation.
Solution
For each machine, consider the net cash flows for all years during the estimated (maximum)
life. Compare them over the LCM of 14 years.
PW 1 12,000 12,000(PF,15%,7) 3000(PA,15%,14) $663
PW 2 8000 1000(PA,15%,5) 3000(PA,15%,9)(PF,15%,5)
$2470
Machine 2 is selected since its PW value is numerically larger than that of machine 1 at 15%.
This result is the opposite of the payback period decision. The PW analysis accounts for the
increased cash flows for machine 2 in the later years. As illustrated in Figure 13–7 (for one life
cycle for each machine), payback analysis neglects all cash flow amounts that may occur after
the payback time has been reached.
Comment
This is a good example of why payback analysis is best used for initial screening and supplemental
risk assessment. Often a shorter-lived alternative evaluated by payback analysis may
appear to be more attractive, when the longer-lived alternative has cash flows later in its life
that make it more economically attractive.
As mentioned in the introduction to this section, breakeven and payback analyses are allied.
They can be used in conjunction to determine the payback period when a desired level of breakeven
is specified. The reverse is also possible; when a desired payback period is established, the
breakeven value with or without a return requirement can be determined. By working together in
this fashion, better economic decisions can be made. Example 13.6 illustrates the second of the
situations mentioned above.

352 Chapter 13 Breakeven and Payback Analysis
EXAMPLE 13.6
The president of a local company expects a product to have a profitable life of between 1 and
5 years. Help her determine the breakeven number of units that must be sold annually (without
any return) to realize payback for each of the time periods 1 year, 2 years, and so on up to
5 years. The cost and revenue estimates are as follows:
Fixed costs: Initial investment of $80,000 with $1000 annual operating cost.
Variable cost: $8 per unit.
Revenue: Twice the variable cost for the first 5 years and 50% of the variable cost thereafter.
Solution by Hand
Define X BE as the breakeven quantity and n p as the payback period. Since values of X BE are
sought for n p 1, 2, 3, 4, 5, solve for breakeven by substituting each payback period. First
develop the FC, r, and v terms.
80,000
Fixed cost, FC
———
n 1000
p
Revenue per unit, r $16 (years 1 through 5 only)
Variable cost per unit, v $8
The breakeven relation from Equation [13.2] is
X BE 80,000n p 1000
————————
[13.10]
8
Insert n p values and solve for X BE , the breakeven value.
n p , payback years 1 2 3 4 5
X BE , units per year 10,125 5125 3458 2625 2125
Solution by Spreadsheet
Figure 13–8 presents a spreadsheet solution for the breakeven values. Equation [13.10] is encoded
to display the answers in column C. The breakeven values are the same as those above,
e.g., sell 5125 units per year to pay back in 2 years. The breakeven curve rapidly flattens out as
shown in the accompanying chart in Figure 13–8.
-
Figure 13–8
Breakeven number of units for different payback periods, Example 13.6.
13.4 More Breakeven and Payback
Analysis on Spreadsheets
The Goal Seek tool that we have used previously is an excellent tool to perform breakeven and payback
analysis. Examples 13.7 and 13.8 demonstrate the use of Goal Seek for both types of problems.

13.4 More Breakeven and Payback Analysis on Spreadsheets 353
EXAMPLE 13.7
The Naruse brake-accelerator pedal (www.autoblog.comtagMasuyukiNaruse) is designed
to minimize the chances that a driver will accidently step on the accelerator pedal of the car
when the brake pedal is the intended target. The design is based on the fact that a person naturally
steps downward on his or her foot when surprised, shocked, or struck with a medical
emergency. In this pedal design, downward motion of the foot will always engage the brake,
never the accelerator. Assume that for the manufacture of pedal components, two equally qualified
machines have been identified and estimates made.
Machine 1 Machine 2
First cost, $ 80,000 110,000
Net cash flow, $ per year 25,000 22,000
Salvage value, $ 2,000 3,000
Life, years 4 6
Using an AW analysis at MARR 10%, the spreadsheet screen shot in Figure 13–9 indicates
that machine 1 is the economic choice with a positive AW value of $193. However, the automated
controls, safety features, and ergonomic design of machine 2 make it a better choice for
the plant in the opinion of the project engineer. Use breakeven analysis to find the threshold
values for each of several parameters that will make machine 2 equally qualified economically.
The parameters to concentrate on are (a) first cost, (b) net cash flow, and (c) life of machine 2,
if all other estimates remain the same.
PMT($B$1,4,NPV($B1,B5:B8)+B4)
Figure 13–9
AW values for two machines, Example 13.7.
Solution
Figure 13–10 shows the spreadsheet and Goal Seek templates that determine breakeven values
for first cost and NCF.
(a) Figure 13–10a: By forcing the AW for machine 2 to equal $193, Goal Seek finds a breakeven
of $96,669. If the first cost can be negotiated down to this cost from $110,000, machine
2 will be economically equivalent to machine 1.
(b) Figure 13–10b: (Remember to reset the first cost to $ –110,000 on the spreadsheet.) By setting
all NCFs equal to the value in year 1 (using the function $C$5), Goal Seek determines
a breakeven of $25,061 per year. Therefore, if the NCF estimate can realistically be
increased from $22,000 to $25,061, again machine 2 will be economically equivalent.
(c) Finding an extended life estimate for machine 2 is a payback question; Goal Seek is not
needed. The easiest approach is to use the NPER function to find the payback period.
Entering NPER(10%,22000,–110000,3000) displays n p 7.13 years. Therefore, extending
the estimated life from 6 to between 7 and 8 years and retaining the salvage value
of $3000 will select machine 2 over 1.

354 Chapter 13 Breakeven and Payback Analysis
(a) First cost
(b) NCF
Figure 13–10
Breakeven values for (a) first cost and (b) annual net cash flow using Goal Seek, Example 13.7.
EXAMPLE 13.8
Chris and her father just purchased a small office building for $160,000 that is in need of a lot
of repairs, but is located in a prime commercial area of the city. The estimated costs each year
for repairs, insurance, etc. are $18,000 the first year, increasing by $1000 per year thereafter.
At an expected 8% per year return, use spreadsheet analysis to determine the payback period
if the building is (a) kept for 2 years and sold for $290,000 sometime beyond year 2 or (b) kept
for 3 years and sold for $370,000 sometime beyond 3 years.
Solution
Figure 13–11 shows the annual costs (column B) and the sales prices if the building is kept 2
or 3 years (columns C and E, respectively). The NPV function is applied (columns D and F) to
determine when the PW changes sign from plus to minus. These results bracket the payback
period for each retention period and sales price. When PW 0, the 8% return is exceeded.
(a) The 8% return payback period is between 3 and 4 years (column D). If the building is sold
after exactly 3 years for $290,000, the payback period was not exceeded; but after 4 years
it is exceeded.
(b) At a sales price of $370,000, the 8% return payback period is between 5 and 6 years (column
F). If the building is sold after 4 or 5 years, the payback is not exceeded; however, a
sale after 6 years is beyond the 8%-return payback period.
If kept 2 years and
sold, payback is
between 3 and 4
NPV(8%,$B$4:B7)+$B$3 PV(8%,A7,,290000)
If kept 3 years and
sold, payback is
between 5 and 6
Figure 13–11
Payback period analysis, Example 13.8

Problems 355
CHAPTER SUMMARY
The breakeven point for a variable for one project is expressed in terms such as units per year or
hours per month. At the breakeven amount Q BE , there is indifference to accept or reject the project.
Use the following decision guideline:
Single Project (Refer to Figure 13–2.)
Estimated quantity is larger than Q BE → accept project
Estimated quantity is smaller than Q BE → reject project
For two or more alternatives, determine the breakeven value of the common variable. Use the
following guideline to select an alternative:
Two Alternatives (Refer to Figure 13–5.)
Estimated level is below breakeven → select alternative with higher
variable cost (larger slope)
Estimated level is above breakeven → select alternative with lower
variable cost (smaller slope)
Payback analysis estimates the number of years necessary to recover the initial investment plus
a stated rate of return. This is a supplemental analysis technique used primarily for initial screening
prior to a full evaluation by PW or some other method. The technique has some drawbacks,
especially for no-return payback analysis, where i 0% is the stated return.
PROBLEMS
Breakeven Analysis for a Project
13.1 A design-to-cost approach to product pricing involves
determining the selling price of the product
and then figuring out if it can be made at a cost lower
than that. Banner Engineering’s QT50R radar-based
sensor features frequency-modulated technology to
accurately monitor or detect objects up to 15 miles
away while resisting rain, wind, humidity, and extreme
temperatures. It has a list price of $589, and
the variable cost of manufacturing the unit is $340.
( a) What could the company’s fixed cost per
year be in order for Banner to break even
with sales of 9000 units per year?
( b) If Banner’s fixed cost is actually $750,000
per year, what is the profit at a sales level of
7000 units per year?
13.2 Handheld fiber-optic meters with white light polarization
interferometry are useful for measuring
temperature, pressure, and strain in electrically
noisy environments. The fixed costs associated
with manufacturing are $800,000 per year. If a
base unit sells for $2950 and its variable cost is
$2075, ( a ) how many units must be sold each year
for breakeven and ( b ) what will the profit be for
sales of 3000 units per year?
13.3 A metallurgical engineer has estimated that the
capital investment cost for recovering valuable
metals (nickel, silver, platinum, gold, etc.) from
the copper refinery’s wastewater stream will be
$12 million. The equipment will have a useful life
of 15 years with no salvage value. Its operating
cost is represented by the relation ($2,600,000) E 1.9 ,
where E is the efficiency of the metal recovery operation
(in decimal form). The amount of metal
currently discharged is 2880 pounds per year prior
to recovery operations, and the efficiency of
recovery is estimated at 71%. What must the
average selling price per pound be for the precious
metals that are recovered and sold in order
for the company to break even at its MARR of
15% per year?
Problems 13.4 through 13.7 are based on the following
information.
Hambry Enterprises produces a component for recycling
uranium used as a nuclear fuel in power plant generators
in France and the United States. Use the following cost
and revenue figures, quoted in U.S. dollars per hundredweight
(cwt), recorded for this year to calculate the answers
for each plant.
Location
Fixed Cost,
$ million
Revenue,
$ per cwt
Cost,
$ per cwt
France 3.50 8,500 3,900
United States 2.65 12,500 9,900

356 Chapter 13 Breakeven and Payback Analysis
13.4 Determine the breakeven point for each plant.
13.5 Estimate the minimum revenue per hundredweight
required for next year if breakeven values and
variable costs remain constant, but fixed costs
increase by 10%.
13.6 During this year, the French plant sold 950 units in
Europe, and the U.S. plant sold 850 units. Determine
the year’s profit (loss) for each plant.
13.7 Hambry’s president has a goal of $1 million profit
next year at each plant with no revenue or fixed
cost increases. Determine the decreases in dollar
amounts and percentages in variable cost necessary
to meet this goal, if the number of units sold is
the same as this year.
13.8 The National Highway Traffic Safety Administration
raised the average fuel efficiency standard to
35.5 miles per gallon (mpg) for cars and light
trucks by the year 2016. The rules will cost consumers
an average of $926 extra per vehicle in the
2016 model year. Assume a person purchases a
new car in 2016 that gets 35.5 mpg and keeps it
for 5 years. If the person drives an average of
1000 miles per month and gets an extra 10 miles
per gallon of gasoline, how much will the gasoline
have to cost in order for the buyer to recover the
extra investment in 5 years at an interest rate of
0.75% per month?
13.9 A call center in India used by U.S. and U.K. credit
card holders has a capacity of 1,400,000 calls annually.
The fixed cost of the center is $775,000
with an average variable cost of $1 and revenue of
$2.50 per call.
(a) Find the percentage of the call capacity that
must be placed each year to break even.
(b) The center manager expects to dedicate the
equivalent of 500,000 of the 1,400,000 capacity
to a new product line. This is expected to
increase the center’s fixed cost to $900,000 of
which 50% will be allocated to the new product
line. Determine the average revenue per
call necessary to make 500,000 calls the
breakeven point for only the new product.
How does this required revenue compare with
the current center revenue of $2.50 per call?
13.10 The addition of a turbocharger to a small V-6 engine
that gets 18 miles per gallon of gasoline can
boost its power to that of a V-8 engine and increase
fuel efficiency at the same time. If Bill will pay
$800 to turbocharge his engine and his fuel efficiency
increases by 3 miles per gallon, how many
miles will he have to drive each month for 3 years
in order to break even? Assume the cost of gasoline
is $3.25 per gallon and the interest rate is 1%
per month.
13.11 Transporting extremely heavy patients (people
who weigh more than 500 pounds) is much more
difficult than transporting normal-weight patients.
Various cities in Colorado, Nebraska, and Kansas
charge $1421 for an extremely obese patient compared
to $758 for a typical patient. The extra fees
are justified by the ambulance companies on the
basis of the specialty equipment required and the
extra personnel involved. If it is assumed that
50 extremely obese patients are transported every
year, how much could the ambulance companies
afford to spend on the specialty equipment now
and break even on the initial cost in 5 years just
from the extra charges? Assume the extra costs are
$400 per patient and the company’s MARR is
10% per year.
13.12 High-profile vehicles have poor fuel efficiency because
of increased wind resistance from their large
front area. A number of companies make devices
that they claim will significantly increase a vehicle’s
fuel efficiency. One company claims that by
spending $560 on its friction-reducing device, the
fuel efficiency of a pickup truck will increase by
25%. Assuming the device works as claimed, for a
vehicle that currently gets 20 miles per gallon
(mpg), how many miles would the owner have to
drive each year to break even in 5 years? Assume
the cost of gasoline is $3.50 per gallon and the
interest rate is 10% per year.
13.13 As the price of gasoline goes up, people are willing
to drive farther to fill their tank in order to save
money. Assume you had been buying gasoline for
$2.90 per gallon and that it went up to $2.98 per
gallon at the station where you usually go. If you
drive an F-150 pickup that gets 18 miles per gallon,
what is the round-trip distance you can drive
to break even if it will take 20 gallons to fill your
tank? Use an interest rate of 8% per year.
13.14 An automobile company is investigating the advisability
of converting a plant that manufactures
economy cars into one that will make retro sports
cars. The initial cost for equipment conversion
will be $200 million with a 20% salvage value
anytime within a 5-year period. The cost of producing
a car will be $21,000, but it is expected to
have a selling price of $33,000 to dealers. The
production capacity for the first year will be
4000 units. At an interest rate of 12% per year, by
what uniform amount will production have to increase
each year in order for the company to recover
its investment in 3 years?

Problems 357
13.15 For the last 2 years, The Health Company has
experienced a fixed cost of $850,000 per year
and an ( r v ) value of $1.25 per unit for its
multivitamin line of products. International
competition has become severe enough that
some financial changes must be made to keep
market share at the current level.
( a) Perform a spreadsheet-based graphical analysis
to estimate the effect on the breakeven
point if the difference between revenue and
variable cost per unit increases somewhere
between 1% and 15% of its current value.
( b) If fixed costs and revenue per unit remain at
their current values, what type of change
must take place to make the breakeven point
go down?
13.16 (This is an extension of Problem 13.15) Expand
the analysis performed in Problem 13.15 by changing
the variable cost per unit. The financial manager
estimates that fixed costs will fall to $750,000
when the required production rate to break even is
at or below 600,000 units. What happens to the
breakeven points over the ( r v ) range of 1% to
15% increase as evaluated previously?
Breakeven Analysis Between Alternatives
13.17 Providing restrooms at parks, zoos, and other cityowned
recreation facilities is a considerable expense
for municipal governments. City councils
usually opt for permanent restrooms in larger
parks and portable restrooms in smaller ones. The
cost of renting and servicing a portable restroom
is $7500 per year. In one northeastern municipality,
the parks director informed the city council
that the cost of constructing a permanent restroom
is $218,000 and the annual cost of maintaining it
is $12,000. He remarked that the rather high cost
is due to the necessity to use expensive materials
and construction techniques that are tailored to
minimize damage from vandalism that often occurs
in unattended public facilities. If the useful
life of a permanent restroom is assumed to be 20
years, how many portable restrooms could the
city afford to rent each year and break even with
the cost of one permanent facility? Let the interest
rate be 6% per year.
13.18 A consulting engineer is considering two methods
for lining ponds used for evaporating concentrate
generated during reverse osmosis treatment of
brackish groundwater for the Clay County Industrial
Park. A geosynthetic bentonite clay liner
(GCL) will cost $1.8 million to install, and if it is
renovated after 4 years at a cost of $375,000, its
life can be extended another 2 years. Alternatively,
a high-density polyethylene (HDPE) geomembrane
can be installed that will have a useful life of
12 years. At an interest rate of 6% per year, how
much money can be spent on the HDPE liner for
the two methods to break even?
13.19 An irrigation canal contractor wants to determine
whether he should purchase a used Caterpillar
mini excavator or a Toro powered rotary tiller for
servicing irrigation ditches in an agricultural area
of California. The initial cost of the excavator is
$26,500 with a $9000 salvage value after 10 years.
Fixed costs for insurance, license, etc. are expected
to be $18,000 per year. The excavator will require
one operator at $15 per hour and maintenance at
$1 per hour. In 1 hour, 0.15 mile of ditch can be
prepared. Alternatively, the contractor can purchase
a tiller and hire 2 workers at $11 per hour
each. The tiller costs $1200 and has a useful life of
5 years with no salvage value. Its operating cost is
expected to be $1.20 per hour, and with the tiller,
the two workers can prepare 0.04 mile of ditch in
1 hour. The contractor’s MARR is 10% per year.
Determine the number of miles of ditch per year
the contractor would have to service for the two
options to break even.
13.20 An effective method to recover water used for regeneration
of ion exchange resins is to use a reverse
osmosis system in a batch treatment mode.
Such a system involves recirculation of the partially
treated water back into the feed tank, causing
the water to heat up. The water can be cooled
using one of two systems: a single-pass heat exchanger
or a closed-loop heat exchange system.
The single-pass system, good for 3 years, requires
a small chiller costing $920 plus stainless steel
tubing, connectors, valves, etc. costing $360. The
cost of water, treatment charges, electricity, etc.
will be $3.10 per hour. The closed-loop system
will cost $3850 to buy, will have a useful life of
5 years, and will cost $1.28 per hour to operate.
What is the minimum number of hours per year
that the cooling system must be used in order to
justify purchase of the closed-loop system? The
MARR is 10% per year, and the salvage values
are negligible.
13.21 Samsung Electronics is trying to reduce supply
chain risk by making more responsible make-buy
decisions through improved cost estimation. A
high-use component (expected usage is 5000 units
per year) can be purchased for $25 per unit with
delivery promised within a week. Alternatively,
Samsung can make the component in-house and
have it readily available at a cost of $5 per unit, if
equipment costing $150,000 is purchased. Labor

358 Chapter 13 Breakeven and Payback Analysis
and other operating costs are estimated to be
$35,000 per year over the study period of 5 years.
Salvage is estimated at 10% of first cost and
i 12% per year. Neglect the element of availability
( a ) to determine the breakeven quantity
and ( b ) to recommend making or buying at the expected
usage level.
13.22 A partner in a medium-size A/E (architecturalengineering)
design firm is evaluating two alternatives
for improving the exterior appearance of the building
they occupy. The building can be completely
painted at a cost of $6500. The paint is expected to
remain attractive for 4 years, at which time repainting
will be necessary. Every time the building is
repainted, the cost will be 20% higher than the previous
time. Alternatively, the building can be sandblasted
now and every 6 years at a cost 40% greater
than the previous time. If the company’s MARR is
10% per year, what is the maximum amount that
could be spent now on the sandblasting alternative
that would render the two alternatives indifferent
over a study period of 12 years?
13.23 A junior mechanical engineering student is cooping
this semester at Regency Aircraft, which
customizes the interiors of private and corporate
jets. Her first assignment is to develop the specifications
for a new machine to cut, shape, and sew
leather or vinyl covers and trims. The first cost is
not easy to estimate due to many options, but the
annual revenue and M&O costs should net out at
$15,000 per year over a 10-year life. Salvage is
expected to be 20% of the first cost. Determine the
breakeven first cost of the machine to just recover
its first cost and a return of 8% per year under two
scenarios:
I: No outside revenue will be developed by
the machine.
II: Outside contracting will occur with estimated
revenue of $10,000 the first year, increasing
by $5000 per year thereafter.
Solve using ( a ) hand and ( b ) spreadsheet solutions.
13.24 Ascarate Fishing Club (a nonprofit organization
dedicated to teaching kids how to fish) is considering
two options for providing a heavily stocked
pond for kids who have never caught a fish before.
Option 1 is an above-ground swimming pool made
of heavy vinyl plastic that will be assembled and
disassembled for each quarterly event. The purchase
price will be $400. Leaks from hooks piercing
the fabric will be repaired with a vinyl repair
kit at a cost of $70 per year, but the pool will have
to be replaced when too many repairs have been
made.
Option 2 is an in-ground pond that will be excavated
by club members at no cost and lined with
fabric that costs $1 per square foot. The pond will
be 15 ft in diameter and 3 ft deep. Assume 300 ft 2
of liner will be purchased. A chain link fence at
$10 per lineal foot will be installed around the
pond (100 ft of fence). Maintenance inside the
fence is expected to cost $20 per year. The park
where the pond will be constructed has committed
the land for only 10 years. At an interest rate of 6%
per year, how long would the above-ground pool
have to last to break even?
13.25 A rural subdivision has several miles of access
roads that need a new surface treatment. Alternative
1 is a gravel base and pavement with an initial
cost of $500,000 that will last for 15 years
and has an annual upkeep cost of $100 per mile.
Alternative 2 is to enhance the gravel base now at
a cost of $50,000 and immediately coat the surface
with a durable hot oil mix, which costs $130
per barrel applied. Annual reapplication of the
mix is required. A barrel covers 0.05 mile. ( a ) If
the discount rate is 6% per year, determine the
number of miles at which the two alternatives
break even. ( b ) A drive in a pickup indicates a
total of 12.5 miles of road. Which is the more
economical alternative?
13.26 A waste-holding lagoon situated near the main
plant receives sludge daily. When the lagoon is
full, it is necessary to remove the sludge to a site
located 8.2 kilometers from the main plant. Currently,
when the lagoon is full, the sludge is removed
by pump into a tank truck and hauled away.
This process requires the use of a portable pump
that initially costs $800 and has an 8-year life. The
company pays a contract individual to operate the
pump and oversee environmental and safety factors
at a rate of $100 per day, plus the truck and
driver must be rented for $200 per day. The company
has the option to install a pump and pipeline
to the remote site. The pump would have an initial
cost of $1600 and a life of 10 years and will cost
$3 per day to operate. The company’s MARR is
10% per year.
(a)
If the pipeline will cost $12 per meter to construct
and will have a 10-year life, how many
days per year must the lagoon require pumping
to justify construction of the pipeline?
(b) If the company expects to pump the lagoon
once per week every week of the year, how
much money can it afford to spend now on
the 10-year life pipeline to just break even?
13.27 Lorraine can select from two nutrient injection
systems for her cottage industry of hydroponic

Problems 359
tomato and lettuce greenhouses. ( a ) Use an AW
relation to determine the minimum number of
hours per year to operate the pumps that will justify
the Auto Green system, if the MARR is 10%
per year. ( b ) Which pump is economically better if
it operates 7 hours per day, 365 days per year?
Nutra Jet (N) Auto Green (A)
Initial cost, $ 4,000 10,300
Life, years 3 6
Rebuild cost, $ 1,000 2,200
Time before rebuild, annually 2,000 8,000
or minimum hours
Cost to operate, $ per hour 1.00 0.90
13.28 An engineering practitioner can lease a fully
equipped computer and color printer system for
$800 per month or purchase one for $8500 now and
pay a $75 per month maintenance fee. If the nominal
interest rate is 15% per year, determine the
months of use necessary for the two to break even.
Show both ( a ) hand and ( b ) spreadsheet solutions.
13.29 The office manager of an environmental engineering
consulting firm was instructed to make an ecofriendly
decision in acquiring an automobile for
general office use. He is considering a gasolineelectric
hybrid or a gasoline-free, all-electric
hatchback. The hybrid under consideration is
GM’s Volt, which will cost $35,000, will have a
salvage value of $15,000 after 5 years, and will
have a range of 40 miles on the electric battery,
plus several hundred more miles when the gasoline
engine kicks in. Nissan’s Leaf, on the other
hand, is a pure electric that will have a range of
only 100 miles, after which its lithium ion battery
must be recharged. The Leaf’s relatively limited
range creates a psychological effect known as
range anxiety (RA), which has the company leaning
toward purchasing the Volt. The Leaf can be
leased for $349 per month after an initial $500
down payment.
The accountant for the consulting firm told the
office manager that the Leaf is the better economic
option based on an evaluation she performed earlier.
If the office manager purchases the Volt anyway
(instead of leasing the Leaf), what is the
monthly equivalent (AW value) of the extra
amount of money the company will be paying to
eliminate range anxiety? Assume the operating
costs will be the same for both vehicles and the
MARR is 0.75% per month.
Payback Analysis
13.30 How long will you have to sell a product that has
an income of $5000 per month and expenses of
$1500 per month if your initial investment is
$28,000 and your MARR is ( a ) 0% and ( b ) 3% per
month? Solve by formula. ( c ) Write the spreadsheet
functions to display the payback period for
both 0% and 3% per month.
13.31 (a) Determine the payback period at an interest
rate of 8% per year for an asset that initially
cost $28,000, has a scrap value of $1500
whenever it is sold, and generates cash flow
of $2900 per year.
(b) If the asset will be in service for 12 years,
should it be purchased?
13.32 ABB purchased fieldbus communication equipment
for a project in South Africa for $3.15 million.
The net cash flow is estimated at $500,000
per year, and a salvage value of $400,000 is anticipated
regardless of when it is sold. Determine the
number of years the equipment must be used to
obtain payback at MARR values of ( a ) 0% and 8%
per year and ( b ) 15% and 16% per year. ( c ) Use a
spreadsheet to plot the payback years for all four
return values.
13.33 Sundance Detective Agency purchased new surveillance
equipment with the following estimates.
The year index is k 1, 2, 3, . . . .
First cost $1050
Annual maintenance cost $70 5 k per year
Extra annual revenue $200 50 k per year
Salvage value
$600 for all years
( a) Calculate the payback period to make a return
of 10% per year.
( b) For a preliminary conclusion, should the
equipment be purchased if the actual useful
life is 7 years?
13.34 Clarisa, an engineering manager, wants to purchase
a resort accommodation to rent to skiers.
She is considering the purchase of a three- bedroom
lodge in upper Montana that will cost $250,000.
The property in the area is rapidly appreciating in
value because people anxious to get away from
urban developments are bidding up the prices. If
Clarisa spends an average of $500 per month for
utilities and the investment increases at a rate of
2% per month, how long would it be before she
could sell the property for $100,000 more than she
has invested in it?
13.35 Laura’s grandparents helped her purchase a small
self-serve laundry business to make extra money
during her 5 college years. When she completed
her engineering management degree, she sold the
business and her grandparents told her to keep the

360 Chapter 13 Breakeven and Payback Analysis
money as a graduation present. For the net cash
flows listed below, determine the following:
(a) The percentage of the investment recovered
during the 5 years
(b) The actual rate of return over the 5-year
period
(c) How long it took to pay back the $75,000 investment
in year 0, plus a 7% per year return
Year 0 1 2 3 4 5
NCF, $ per year 75,000 10,500 18,600 2000 28,000 105,000
13.36 Buhler Tractor sold a tractor for $45,000 to Tom
Edwards 10 years ago. ( a ) What is the uniform net
cash flow that Tom must make each year to realize
payback and a return of 5% per year on his investment
over a period of 3 years? 5 years? 8 years?
All 10 years? ( b ) If the net cash flow was actually
$5000 per year, what is the amount Tom should
have paid for the tractor to realize payback plus the
5% per year return over these 10 years?
13.37 National Parcel Service has historically owned and
maintained its own delivery trucks. Leasing is an
option being seriously considered because costs
for maintenance, fuel, insurance, and some liability
issues will be transferred to Pacific Leasing, the
truck leasing company. The study period is no
more than 24 months for either alternative. The annual
lease cost is paid at the beginning of each year
and is not refundable for partially used years. Use
the first cost and net cash flow estimates to determine
the payback in months with a nominal 9%
per year return for the ( a ) purchase option and
( b ) lease option.
Purchase:
Lease:
P $– 30,000 now
Monthly cost $1000
Monthly revenue $4500
P $10,000 at the beginning
of each year (months 0 and 12)
Monthly cost $−2500
Monthly revenue $4500
13.38 Julian Browne, owner of Clear Interior Environments,
purchased an air scrubber, HEPA vacuum,
and other equipment for mold removal for $15,000
eight months ago. Net cash flows were $−2000 for
each of the first 2 months, followed by $1000 per
month for months 3 and 4. For the last 4 months, a
contract generated a net $6000 per month. Julian
sold the equipment yesterday for $3000 to a friend.
Determine ( a ) the no-return payback period and
( b ) the nominal 18%-per-year payback period.
13.39 Explain why payback analysis may favor an alternative
with a shorter payback period when it is not
the better choice economically.
13.40 When comparing two alternatives, why is it best to
use no-return payback analysis as a preliminary
screening tool prior to conducting a complete PW
or AW evaluation?
Spreadsheet Problems
13.41 Benjamin used regression analysis to fit quadratic
relations to monthly revenue and cost data with the
following results:
R 0.007 Q 2 32 Q
TC 0.004 Q 2 2.2 Q 8
( a) Plot R and TC. Estimate the quantity Q p at
which the maximum profit should occur.
Estimate the amount of profit at this quantity.
(b) The profit relation P R TC and calculus
can be used to determine the quantity Q p at
which the maximum profit will occur and the
amount of the profit. The equations are
Profit aQ 2 bQ c
Q p —— b
2a
Maximum profit b2 ——
4a c
Use these relations to confirm the graphical
estimates you made in ( a ). (Your instructor
may ask you to derive the relations above.)
13.42 The National Potato Cooperative purchased a deskinning
machine last year for $150,000. Revenue
for the first year was $50,000. Over the total estimated
life of 8 years, what must the remaining
equivalent annual revenues (years 2 through 8)
equal to break even by recovering the investment
and a return of 10% per year? Costs are expected
to be constant at $42,000 per year, and a salvage
value of $20,000 is anticipated.
Problems 13.43 and 13.44 are based on the following
information.
Wilson Partners manufactures thermocouples for electronics
applications. The current system has a fixed cost of $300,000
per year and a variable cost of $10 per unit. Wilson sells the
units for $14 each. A newly proposed process will add onboard
features that allow the revenue to increase to $16 per
unit, but the fixed cost will now be $500,000 per year. The
variable cost of the new system will be based on a $48 per
hour rate with 0.2 hour required to produce each unit.
13.43 Determine the annual breakeven quantity for
( a ) the current system and ( b ) the new system.
13.44 Plot the two profit relations and estimate graphically
the breakeven quantity between the two
alternatives.

Additional Problems and FE Exam Review Questions 361
Problems 13.45 through 13.48 are based on the following
information.
Mid-Valley Industrial Extension Service, a state- sponsored
agency, provides water quality sampling services to all
business and industrial firms in a 10-county region. Just
last month, the service purchased all necessary lab equipment
for full in-house testing and analysis. Now an outsourcing
agency has offered to take over this function on a
per sample basis. Data and quotes for the two options have
been collected. The MARR for government projects is 5%
per year, and a study period of 8 years is chosen.
In-house: Equipment and supplies initially cost
$125,000 for a life of 8 years, an AOC of
$15,000, and annual salaries of $175,000.
Sample costs average $25 each. There is
no significant market value for the equipment
and supplies currently owned.
Outsourced: Contractors quote sample cost averages
of $100 for the first 5 years, increasing to
$125 per sample for years 6 through 8.
13.45 Determine the breakeven number of tests between
the two options.
13.46 Use a spreadsheet to graph the AW curves for both
options for test loads between 0 and 4000 per year
in increments of 1000 tests. What is the estimated
breakeven quantity?
13.47 The service director has asked the outsource company
to reduce the per sample costs by 25% across
the board over the 8-year study period. What will this
do to the breakeven point? ( Hint : Look carefully at
your graph from Problem 13.46 before answering.)
13.48 Assume the Extension Service can reduce its annual
salaries from $175,000 to $100,000 per
year and the per sample cost from $25 to $20.
What will this do to the breakeven point? ( Hint :
Again, look carefully at your graph from the
previous problem before answering.) What is
the new annual breakeven test quantity?
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
13.49 In linear breakeven analysis, if a company expects
to operate at a point above the breakeven point, it
should select the alternative:
( a) With the lower fixed cost
( b) With the higher fixed cost
( c) With the lower variable cost
( d) With the higher variable cost
13.50 A company is considering two alternatives to automate
the pH of process liquids. Alternative A will
have fixed costs of $42,000 per year and will require
2 workers at $48 per day each. Together,
these workers can generate 100 units of product
per day. Alternative B will have fixed costs of
$56,000 per year, but with this alternative, 3 workers
will generate 200 units of product. If x is the
number of units per year, the variable cost (VC) in
$ per year for alternative B is represented by:
( a) [2(48)100] x
( b) [3(48)200] x
( c) [3(48)200] x 56,000
( d) [2(48)100] x 42,000
13.52 AW 1 23,000( AP,10%,10) 4000( A F ,10%,10)
3000 3 x
AW 2 8,000(AP,10%,4) 2000 − 6 x
For these two AW relations, the breakeven point x ,
in miles per year, is closest to:
( a ) 1130
( b ) 1224
( c ) 1590
( d ) 655
13.53 To make an item in-house, equipment costing
$250,000 must be purchased. It will have a life of
4 years, an annual cost of $80,000, and each unit
will cost $40 to manufacture. Buying the item externally
will cost $100 per unit. At i 15% per
year, it is cheaper to make the item in-house if the
number per year needed is:
( a) Above 1047 units
( b) Above 2793 units
( c) Equal to 2793 units
( d) Below 2793 units
13.51 When the variable cost is reduced for linear total
cost and revenue lines, the breakeven point decreases.
This is an economic advantage because:
( a) The revenue per unit will increase.
( b) The two lines will now cross at zero.
( c) The profit will increase for the same revenue
per unit.
( d) The total cost line becomes nonlinear.
13.54 A procedure at Mercy Hospital has fixed costs of
$10,000 per year and variable costs of $50 per test.
If the procedure is automated, its fixed cost will be
$21,500 per year, but its variable cost will be only
$10 per test. The number of tests that must be performed
each year for the two operations to break
even is closest to:
( a ) 290 ( b ) 455 ( c ) 750 ( d ) Over 800

362 Chapter 13 Breakeven and Payback Analysis
13.55 An assembly process can be completed using
either alternative X or Y. Alternative X has fixed
costs of $10,000 per year with a variable cost of
$50 per unit. If the process is automated per alternative
Y, its fixed cost will be $5000 per year
and its variable cost will be only $10 per unit.
The number of units that must be produced each
year in order for alternative Y to be favored is
closest to:
(a) Y will be favored for any level of production
(b) 125
(c) 375
(d) X will be favored for any level of production
13.56 Two different methods are under consideration
for building a bypass road. Material C will cost
$100,000 per mile and last for 10 years. Its annual
maintenance cost will be $10,000 per year
per mile. Material D will cost $30,000 per mile
and last for 5 years. At an interest rate of 6% per
year, the annual maintenance cost for material D
that will make the two methods cost the same is
closest to:
( a ) Less than $14,000 ( b ) $14,270
( c ) $16,470 ( d ) $19,510
13.57 A construction company can purchase a piece of
equipment for $50,000 and spend $100 per day in
operating costs. The equipment will have a 5-year
life with no salvage value. Alternatively, the company
can lease the equipment for $400 per day.
The number of days per year the company must
require the equipment to justify its purchase at an
interest rate of 8% per year is closest to:
( a ) 10 days ( b ) 42 days
( c ) 51 days ( d ) 68 days
13.58 A tractor has a first cost of $40,000, a monthly operating
cost of $1500, and a salvage value of
$12,000 in 10 years. The MARR is 12% per year.
An identical tractor can be rented for $3200 per
month (operating cost not included). If n is the
minimum number of months per year the tractor
must be used in order to justify its purchase, the
relation to find n is represented by:
(a) 40,000( A P ,1%,10) 1500 n
12,000( A F ,1%,10) 3200 n
(b) 40,000( A P ,12%,10) 1500 n
12,000( A F ,12%,10) 3200 n
(c) 40,000( A P ,1%,120) 1500 n
12,000( A F ,1%,120) 3200 n
(d) 40,000( A P ,11.4%,10) 1500 n
12,000( A F ,11.4%,10) 3200 n
13.59 An anticorrosive coating for a chemical storage
tank will cost $5000 and last 5 years if touched
up at the end of 3 years at a cost of $1000. If an
oil-base enamel coating could be used that will
last 2 years, the amount the enamel coating can
cost for the two to breakeven at i 8% per year
is closest to:
( a ) $2120 ( b ) $2390
( c ) $2590 ( d ) $2725
13.60 The price of a car is $50,000 today. Its price is expected
to increase by $2400 each year. You now
have $25,000 in an investment that is earning 20%
per year. The number of years before you have
enough money to buy the car, without borrowing
any money, is closest to:
( a ) 3 years ( b ) 5 years
( c ) 7 years ( d ) 9 years
13.61 Process A has a fixed cost of $16,000 per year and
a variable cost of $40 per unit. For process B,
5 units can be produced in 1 day at a cost of $125.
If the company’s MARR is 10% per year, the fixed
cost of process B that will make the two alternatives
have the same annual cost at a production
rate of 1000 units per year is closest to:
(a) Less than $10,000
(b) $18,000
(c) $27,000
(d) Over $30,000
13.62 The profit relation for the following estimates at a
quantity that is 20% above breakeven is:
Fixed cost $500,000 per year
Variable cost per unit $200
Revenue per unit $250
(a) Profit 200(12,000) 250(12,000) 500,000
(b) Profit 250(12,000) 500,000 200 (12,000)
(c) Profit 250(12,000) 200(12,000) 500,000
(d) Profit 250(10,000) 200(10,000) 500,000
13.63 Two methods of weed control in an irrigation canal
are under consideration. Method A involves lining
at a cost of $4000. The lining will last 20 years.
The maintenance cost with this method will be
$3 per mile per year. Method B involves spraying
a chemical that costs $40 per gallon. One gallon
will treat 8 miles, but the treatment must be applied
4 times per year. In determining the number of
miles per year that would result in breakeven, the
variable cost for method B is closest to:
( a ) $5 per mile ( b ) $15 per mile
( c ) $20 per mile ( d ) $40 per mile
13.64 How long will you have to maintain a business that
has an income of $5000 per year and expenses of
$1500 per year if your initial investment was
$28,000 and your MARR is 10% per year?
( a ) Less than 6 years ( b ) 8 years
( c ) 12 years ( d ) 17 years

Case Study 363
CASE STUDY
WATER TREATMENT PLANT PROCESS COSTS
Background
Aeration and sludge recirculation have been practiced for
many years at municipal and industrial water treatment
plants. Aeration is used primarily for the physical removal of
gases or volatile compounds, while sludge recirculation can
be beneficial for turbidity removal and hardness reduction.
When the advantages of aeration and sludge recirculation
in water treatment were first recognized, energy costs were so
low that such considerations were seldom of concern in treatment
plant design and operation. With the huge increases in
electricity cost that have occurred in some localities, however,
it became necessary to review the cost-effectiveness of
all water treatment processes that consume significant
amounts of energy. This study was conducted at a municipal
water treatment plant for evaluating the cost-effectiveness of
the pre-aeration and sludge recirculation practices.
Information
This study was conducted at a 106 m 3 per minute watertreatment
plant where, under normal operating circumstances,
sludge from the secondary clarifiers is returned to
the aerator and subsequently removed in the primary clarifiers.
Figure 13–12 is a schematic of the process.
To evaluate the effect of sludge recirculation, the sludge
pump was turned off, but aeration was continued. Next, the
sludge pump was turned back on, and aeration was discontinued.
Finally, both processes were discontinued. Results obtained
during the test periods were averaged and compared to
the values obtained when both processes were operational.
The results obtained from the four operating modes
showed that the hardness decreased by 4.7% when both processes
were in operation (i.e., sludge recirculation and aeration).
When only sludge was recirculated, the reduction was
3.8%. There was no reduction due to aeration only, or when
there was neither aeration nor recirculation. For turbidity, the
reduction was 28% when both recirculation and aeration
were used. The reduction was 18% when neither aeration nor
recirculation was used. The reduction was also 18% when
aeration alone was used, which means that aeration alone was
of no benefit for turbidity reduction. With sludge recirculation
alone, the turbidity reduction was only 6%, meaning that
sludge recirculation alone actually resulted in an increase in
turbidity—the difference between 18% and 6%.
Since aeration and sludge recirculation did cause readily
identifiable effects on treated water quality (some good and
others bad), the cost-effectiveness of each process for turbidity
and hardness reduction was investigated. The calculations
are based on the following data:
Aerator motor 40 hp
Aerator motor efficiency 90%
Sludge recirculation motor 5 hp
Recirculation pump efficiency 90%
Electricity cost 9 ¢kWh (previous analysis)
Lime cost 7.9 ¢kg
Lime required 0.62 mgL per mgL hardness
Coagulant cost 16.5 ¢kg
Daysmonth 30.5
As a first step, the costs associated with aeration and sludge
recirculation were calculated. In each case, costs are independent
of flow rate.
Aeration cost:
40 hp 0.75 kWhp 0.09 $kWh 24 hday
0.90 $72 per day or $2196 per month
Sludge recirculation cost:
5 hp 0.75 kWhp 0.09 $kWh 24 hday
0.90 $9 per day or $275 per month
Chemical
additions
Flash
mix
Flocculation
Figure 13–12
Schematic of water
treatment plant.
Primary
clarifier
Secondary
clarifier
Aerator
Filter
To clear
well
Canal

364 Chapter 13 Breakeven and Payback Analysis
TABLE 13–1
Cost Summary in Dollars per Month
Alt.
I.D.
Alternative
Description
Savings from
Discontinuation of
Aeration
(1)
Recirculation
(2)
Total Savings
(3) (1) (2)
Extra Cost for
Removal of
Hardness
(4)
Turbidity
(5)
Total
Extra Cost
(6) (4) (5)
Net
Savings
(7) (3) (6)
1 Sludge recirculation
Normal operating condition
and aeration
2 Aeration only — 275 275 1380 469 1849 1574
3 Sludge recirculation 2196 — 2196 262 845 1107 1089
only
4 Neither aeration nor
sludge recirculation
2196 275 2471 1380 469 1849 622
The estimates appear in columns 1 and 2 of the cost summary
in Table 13–1.
Costs associated with turbidity and hardness removal are
a function of the chemical dosage required and the water flow
rate. The calculations below are based on a design flow of
53 m 3 minute.
As stated earlier, there was less turbidity reduction
through the primary clarifier without aeration than there was
with it (28% versus 6%). The extra turbidity reaching the
flocculators could require further additions of the coagulating
chemical. If it is assumed that, as a worst case, these chemical
additions would be proportional to the extra turbidity, then
22 percent more coagulant would be required. Since the average
dosage before discontinuation of aeration was 10 mgL,
the incremental chemical cost incurred because of the increased
turbidity in the clarifier effluent would be
(10 0.22) mgL 10 –6 kgmg 53 m 3 min
1000 Lm 3 0.165 $kg 60 minh
24 hday $27.70day or $845month
Similar calculations for the other operating conditions (i.e.,
aeration only, and neither aeration nor sludge recirculation)
reveal that the additional cost for turbidity removal would be
$469 per month in each case, as shown in column 5 of
Table 13–1.
Changes in hardness affect chemical costs by virtue of the
direct effect on the amount of lime required for water softening.
With aeration and sludge recirculation, the average hardness
reduction was 12.1 mgL (that is, 258 mgL 4.7%).
However, with sludge recirculation only, the reduction was
9.8 mgL, resulting in a difference of 2.3 mgL attributed to
aeration. The extra cost of lime incurred because of the discontinuation
of aeration, therefore, was
2.3 mgL 0.62 mgL lime 10 6 kgmg
53m 3 min 1000 Lm 3 0.079 $kg
60 minh 24 hday $8.60day or
$262month
When sludge recirculation was discontinued, there was no
hardness reduction through the clarifier, so that the extra lime
cost would be $1380 per month.
The total savings and total costs associated with changes
in plant operating conditions are tabulated in columns 3 and
6 of Table 13–1, respectively, with the net savings shown in
column 7. Obviously, the optimum condition is represented
by “sludge recirculation only.” This condition would result in
a net savings of $1089 per month, compared to a net savings
of $622 per month when both processes are discontinued and
a net cost of $1574 per month for aeration only. Since the
calculations made here represent worst-case conditions, the
actual savings that resulted from modifying the plant operating
procedures were greater than those indicated.
In summary, the commonly applied water treatment practices
of sludge recirculation and aeration can significantly affect
the removal of some compounds in the primary clarifier.
However, increasing energy and chemical costs warrant continued
investigations on a case-by-case basis of the costeffectiveness
of such practices.
Case Study Exercises
1. What will be the monthly savings in electricity from
discontinuation of aeration if the cost of electricity is
now 12 ¢kWh?
2. Does a decrease in the efficiency of the aerator motor
make the selected alternative of sludge recirculation only
more attractive, less attractive, or the same as before?
3. If the cost of lime were to increase by 50%, would the
cost difference between the best alternative and secondbest
alternative increase, decrease, or remain the same?
4. If the efficiency of the sludge recirculation pump were
reduced from 90% to 70%, would the net savings difference
between alternatives 3 and 4 increase, decrease, or
stay the same?
5. If hardness removal were to be discontinued at the treatment
plant, which alternative would be the most costeffective?
6. If the cost of electricity decreased to 8 ¢kWh, which
alternative would be the most cost-effective?
7. At what electricity cost would the following alternatives
just break even? ( a ) Alternatives 1 and 2, ( b ) alternatives
1 and 3, ( c ) alternatives 1 and 4.

LEARNING STAGE 4
Rounding Out the Study
LEARNING STAGE 4
Rounding Out
the Study
CHAPTER 14
Effects of Inflation
CHAPTER 15
Cost Estimation and
Indirect Cost
Allocation
CHAPTER 16
Depreciation Methods
CHAPTER 17
After-Tax Economic
Analysis
This stage includes topics to enhance your ability to perform a
thorough engineering economic study of one project or several
alternatives. The effects of inflation, depreciation,
income taxes in all types of studies, and indirect costs are incorporated
into the methods of previous chapters. Techniques of cost
estimation to better predict cash flows are treated in order to base
alternative selection on more accurate estimates. The last two chapters
include additional material on the use of engineering economics
in decision making. An expanded version of sensitivity analysis is
developed to examine parameters that vary over a predictable range
of values. The use of decision trees and an introduction to real
options are included. Finally, the elements of risk and probability
are explicitly considered using expected values, probabilistic analysis,
and spreadsheet-based Monte Carlo simulation.
Several of these topics can be covered earlier in the text, depending
on the objectives of the course. Use the chart in the Preface to
determine appropriate points at which to introduce the material in
Learning Stage 4.
CHAPTER 18
Sensitivity Analysis
and Staged Decisions
CHAPTER 19
More on Variation
and Decision Making
under Risk

CHAPTER 14
Effects of
Inflation
LEARNING OUTCOMES
Purpose: Consider the effects of inflation when performing an engineering economy evaluation.
SECTION TOPIC LEARNING OUTCOME
14.1 Inflationary impact • Demonstrate the difference that inflation makes
on money now and money in the future; also,
explain deflation.
14.2 PW with inflation • Calculate the PW of cash flows with an
adjustment made for inflation.
14.3 FW with inflation • Determine the real interest rate and calculate
the inflation-adjusted FW with different
interpretations of future worth values.
14.4 CR with inflation • Calculate capital recovery of an investment
using the AW value with inflation considered.

T
his chapter concentrates upon understanding and calculating the effects of inflation
in time value of money computations. Inflation is a reality that we deal with
nearly everyday in our professional and personal lives.
The annual inflation rate is closely watched and historically analyzed by government
units, businesses, and industrial corporations. An engineering economy study can have different
outcomes in an environment in which inflation is a serious concern compared to one
in which it is of minor consideration. In the first decade of the 21st century, inflation has not
been a major concern in the United States or most industrialized nations. But the inflation
rate is sensitive to real, as well as perceived, factors of the economy. Factors such as the cost
of energy, interest rates, availability and cost of skilled people, scarcity of materials, political
stability, and other, less tangible factors have short-term and long-term impacts on the
inflation rate. In some industries, it is vital that the effects of inflation be integrated into an
economic analysis. The basic techniques to do so are covered here.
14.1 Understanding the Impact of Inflation
We are all very well aware that $20 now does not purchase the same amount as $20 did in 2005
and purchases significantly less than in 2000. Why? Primarily this is due to inflation and the
purchasing power of money.
Inflation is an increase in the amount of money necessary to obtain the same amount of
goods or services before the inflated price was present.
Purchasing power , or buying power , measures the value of a currency in terms of the quantity
and quality of goods or services that one unit of money will purchase. Inflation decreases
the purchasing ability of money in that less goods or services can be purchased for the same
one unit of money.
Inflation
Inflation occurs because the value of the currency has changed—it has gone down in value. The
value of money has decreased, and as a result, it takes more money for the same amount of
goods or services. This is a sign of inflation . To make comparisons between monetary amounts
that occur in different time periods, the different-valued money first must be converted to constant-value
money in order to represent the same purchasing power over time . This is especially
important when future sums of money are considered, as is the case with all alternative
evaluations.
Money in one period of time t 1 can be brought to the same value as money in another period
of time t 2 by using the equation
amount in period t
Amount in period t 1 ————————————— 2
[14.1]
inflation rate between t 1 and t 2
Using dollars as the currency, dollars in period t 1 are called constant-value dollars or today’s
dollars. Dollars in period t 2 are called future dollars or then-current dollars and have inflation
taken into account. If f represents the inflation rate per period (year) and n is the number of time
periods (years) between t 1 and t 2 , Equation [14.1] is
Constant-value dollars ——————
future dollars
(1 f ) n [14.2]
Future dollars constant-value dollars(1 f ) n [14.3]
We can express future dollars in terms of constant-value dollars, and vice versa, by applying the
last two equations. This is how the Consumer Price Index (CPI) and cost estimation indices
(of Chapter 15) are determined. As an illustration, use the price of a cheese pizza.
$8.99 March 2011
If inflation on food prices averaged 5% during the last year, in constant-value 2010 dollars, this
cost is last year’s equivalent of
$8.991.05 $8.56 March 2010

368 Chapter 14 Effects of Inflation
A predicted price in 2012, according to Equation [14.3], is
$8.99(1.05) $9.44 March 2012
The price of $9.44 in 2012 buys exactly the same cheese pizza as $8.56 did in 2010. If inflation
averages 5% per year over the next 10 years, Equation [14.3] is used to predict a price in 2020
based on 2010.
$8.56(1.05) 10 $13.94 March 2020
This is a 63% increase over the 2010 price at 5% inflation for prepared food prices, which is generally
not considered excessive. In some areas of the world, hyperinflation may average 50% per
year. In such an unfortunate economy, the cheese pizza in 10 years rises from the dollar equivalent
of $8.99 to $518.44! This is why countries experiencing hyperinflation must devalue the
currency by factors of 100 and 1000 when unacceptable inflation rates persist.
Placed into an industrial or business context, at a reasonably low inflation rate averaging 4%
per year, equipment or services with a first cost of $209,000 will increase by 48% to $309,000
over a 10-year span. This is before any consideration of the rate of return requirement is placed
upon the equipment’s revenue-generating ability. Make no mistake: Inflation is a formidable
force in our economy .
There are three different rates that are important to understanding inflation: the real interest
rate ( i ), the market interest rate ( i f ), and the inflation rate ( f ). Only the first two are interest rates.
Real or inflation-free interest rate i . This is the rate at which interest is earned when the
effects of changes in the value of currency (inflation) have been removed. Thus, the real interest
rate presents an actual gain in purchasing power. (The equation used to calculate i , with the
influence of inflation removed, is derived later in Section 14.3.) The real rate of return that
generally applies for individuals is approximately 3.5% per year. This is the “safe investment”
rate. The required real rate for corporations (and many individuals) is set above this safe rate
when a MARR is established without an adjustment for inflation.
Inflation-adjusted or market interest rate i f . As its name implies, this is the interest rate that
has been adjusted to take inflation into account. This is the interest rate we hear everyday. It is
a combination of the real interest rate i and the inflation rate f , and, therefore, it changes as the
inflation rate changes. It is also known as the infl ated interest rate. A company’s MARR adjusted
for inflation is referred to as the inflation-adjusted or market MARR. The determination
of this value is discussed in Section 14.3.
Inflation rate f . As described above, this is a measure of the rate of change in the value of the
currency.
Deflation is the opposite of inflation in that when deflation is present, the purchasing
power of the monetary unit is greater in the future than at present. That is, it will take fewer
dollars in the future to buy the same amount of goods or services as it does today. Inflation
occurs much more commonly than deflation, especially at the national economy level. In
deflationary economic conditions, the market interest rate is always less than the real interest
rate.
Temporary price deflation may occur in specific sectors of the economy due to the introduction
of improved products, cheaper technology, or imported materials or products that
force current prices down. In normal situations, prices equalize at a competitive level after a
short time. However, deflation over a short time in a specific sector of an economy can be
orchestrated through dumping . An example of dumping may be the importation of materials,
such as steel, cement, or cars, into one country from international competitors at very
low prices compared to current market prices in the targeted country. The prices will go
down for the consumer, thus forcing domestic manufacturers to reduce their prices in order
to compete for business. If domestic manufacturers are not in good financial condition, they
may fail, and the imported items replace the domestic supply. Prices may then return to normal
levels and, in fact, become inflated over time, if competition has been significantly
reduced.
On the surface, having a moderate rate of deflation sounds good when inflation has been
present in the economy over long periods. However, if deflation occurs at a more general

14.2 Present Worth Calculations Adjusted for Inflation 369
level, say nationally, it is likely to be accompanied by the lack of money for new capital. Another
result is that individuals and families have less money to spend due to fewer jobs, less
credit, and fewer loans available; an overall “tighter” money situation prevails. As money
gets tighter, less is available to be committed to industrial growth and capital investment. In
the extreme case, this can evolve over time into a deflationary spiral that disrupts the entire
economy. This has happened on occasion, notably in the United States during the Great Depression
of the 1930s.
Engineering economy computations that consider deflation use the same relations as those for
inflation. For basic equivalence between constant-value dollars and future dollars, Equations
[14.2] and [14.3] are used, except the deflation rate is a − f value. For example, if deflation is estimated
to be 2% per year, an asset that costs $10,000 today would have a first cost 5 years from
now determined by Equation [14.3].
10,000(1 f )
n
10,000(0.98) 5 10,000(0.9039) $9039
14.2 Present Worth Calculations
Adjusted for Inflation
When the dollar amounts in different time periods are to be expressed in constant-value dollars,
the equivalent present and future amounts must be determined using the real interest
rate i . The calculations involved in this procedure are illustrated in Table14–1, where the inflation
rate is 4% per year. Column 2 shows the inflation-driven increase for each of the next
4 years for an item that has a cost of $5000 today. Column 3 shows the cost in future dollars,
and column 4 verifies the cost in constant-value dollars via Equation [14.2]. When the future
dollars of column 3 are converted to constant-value dollars (column 4), the cost is always
$5000, the same as the cost at the start. This is predictably true when the costs are increasing
by an amount exactly equal to the inflation rate. The actual cost (in inflated dollars) of the
item 4 years from now will be $5849, but in constant-value dollars the cost in 4 years will still
amount to $5000. Column 5 shows the present worth of future amounts of $5000 at a real
interest rate of i 10% per year.
Two conclusions can be drawn. At f 4%, $5000 today inflates to $5849 in 4 years. And
$5000 four years in the future has a PW of only $3415 now in constant-value dollars at a real
interest rate of 10% per year.
Figure 14–1 graphs the differences over a 4-year period of the constant-value amount of
$5000, the future-dollar costs at 4% inflation, and the present worth at 10% real interest with
inflation considered. The effect of compounded inflation and interest rates can be large, as you
can see by the shaded area.
An alternative, less complicated method of accounting for inflation in a present worth analysis
involves adjusting the interest formulas themselves to account for inflation. Consider the P F
formula, where i is the real interest rate.
P F ———— 1
(1 i) n
TABLE 14–1 Inflation Calculations Using Constant-Value Dollars (f 4%, i 10%)
Year
n
(1)
Cost
Increase
due to 4%
Inflation, $
(2)
Cost in
Future
Dollars, $
(3)
Future Cost in
Present Worth
Constant-Value
at Real
Dollars, $
i 10%,$
(4) (3)1.04 n (5) (4)(PF,10%,n)
0 5000 5000 5000
1 5000(0.04) 200 5200 5200(1.04) 1 5000 4545
2 5200(0.04) 208 5408 5408(1.04) 2 5000 4132
3 5408(0.04) 216 5624 5624(1.04) 3 5000 3757
4 5624(0.04) 225 5849 5849(1.04) 4 5000 3415

370 Chapter 14 Effects of Inflation
Figure 14–1
Comparison of constantvalue
dollars, future dollars,
and their present
worth values.
6000
Actual cost
at 4% inflation
$5849
Increased
cost
5000
Future dollars
Constant value
10% Present worth
$5000
Decrease in
PW due
to inflation
and interest
4000
$3415
3000
0
1
2
Time, years
3
4
The F , which is a future-dollar amount with inflation built in, can be converted to constant-value
dollars by using Equation [14.2].
P ———— F 1
(1 f ) n ————
(1 i) n
F ——————— 1
(1 i f if ) n [14.4]
If the term i f if is defined as i f , the equation becomes
P F
1 ————
(1 i f ) n F(PF,i f ,n) [14.5]
As described earlier, i f is the inflation-adjusted or market interest rate and is defined as
i f i f if [14.6]
where i real interest rate
f inflation rate
For a real interest rate of 10% per year and an inflation rate of 4% per year, Equation [14.6] yields
a market interest rate of 14.4%.
i f 0.10 0.04 0.10(0.04) 0.144
Table 14–2 illustrates the use of i f 14.4% in PW calculations for $5000 now, which inflates
to $5849 in future dollars 4 years hence. As shown in column 4, the present worth for each year
is the same as column 5 of Table 14–1.
The present worth of any series of cash flows—uniform, arithmetic gradient, or geometric
gradient—can be found similarly. That is, either i or i f is introduced into the P A, P G, or P g factors,
depending upon whether the cash flow is expressed in constant-value (today’s) dollars or
future dollars, respectively.
If a cash flow series is expressed in today’s (constant-value) dollars, then its PW is the discounted
value using the real interest rate i .
If the cash flow is expressed in future dollars, the PW value is obtained using i f .
It is always acceptable to first convert all future dollars to constant-value dollars using
Equation [14.2] and then find the PW at the real interest rate i .

14.2 Present Worth Calculations Adjusted for Inflation 371
TABLE 14–2
Present Worth Calculation Using an Inflated Interest Rate
Year
n
(1)
Cost in
Future Dollars, $
(2)
(PF,14.4%,n)
(3)
PW, $
(4) (2)(3)
0 5000 1 5000
1 5200 0.8741 4545
2 5408 0.7641 4132
3 5624 0.6679 3757
4 5849 0.5838 3415
EXAMPLE 14.1
Glyphosate is the active ingredient in the herbicide Roundup ® marketed by Monsanto Co.
Roundup has been a dependable product used by farmers, municipalities, and suburbanites
alike to control weeds in fields, yards, gardens, streets, and parks. Contributions to Monsanto’s
revenue have been reduced significantly by international dumping of generic glyphosate, as
announced in mid-2010. 1 Monsanto’s sales price was decreased from $16 to $12 per gallon to
compete with the highly competitive pricing, and it is expected that the international price
will settle at approximately $10 per gallon. Assume when the price was set at $16 per gallon,
there was a prediction that in 5 years the price would inflate to $19 per gallon. Perform the
following analysis.
(a) Determine the annual rate of inflation over 5 years to increase the price from $16 to $19.
(b) Using the same annual rate determined above as the rate at which the price continues to
decline from the new $12 price, calculate the expected price in 5 years. Compare this result
with $10 per gallon that Monsanto predicted would be the longer-term price.
(c) Provided Monsanto were somehow able to recover the same market share as it had previously,
and the same inflation rate was applied to the reduced $12 per gallon price, determine
the price 5 years in the future and compare it with the pre-dumping price of $16 per
gallon.
(d) Determine the market interest rate that must be used in economic equivalence
computations, if inflation is considered and an 8% per year real return is expected
by Monsanto.
Solution
The first three parts involve inflation only—no return on investments.
(a) Solve Equation [14.2] for the annual inflation rate f with known constant-value and future
amounts.
16 19(PF,f,5) ———— 19
(1 f ) 5
1 f (1.1875) 0.2
f 0.035 (3.5% per year)
(b) If the price deflation rate is 3.5% per year, find the F value in 5 years with P $12.
F P(FP,3.5%,5) 12(1 0.035) 5
12(0.8368)
$10.04
The price will fall to exactly $10 per gallon after 5 years, as Monsanto predicted.
1 S. Kilman and I. Berry, “Monsanto Cuts Roundup Prices as Knockoffs Flood Farm Belt,” Wall Street
Journal, May 28, 2010.

372 Chapter 14 Effects of Inflation
(c) Five years in the future, at 3.5% per year inflation, the price will be
F P(FP,3.5%,5) 12(1.035) 5
12(1.1877)
$14.25
After 5 years of recovery at the same level as historically experienced, the price will still
be considerably lower than it was at the pre-dumping point ($14.25 versus $16 per gallon).
(d) With inflation at 3.5% per year and a real return of 8% per year, Equation [14.6] results in
a market rate of 11.78% per year.
i f 0.08 0.035 (0.08)(0.035)
0.1178 (11.78% per year)
EXAMPLE 14.2
A 15-year $50,000 bond that has a dividend rate of 10% per year, payable semiannually, is currently
for sale. If the expected rate of return of the purchaser is 8% per year, compounded
semiannually, and if the inflation rate is expected to be 2.5% each 6-month period, what is the
bond worth now (a) without an adjustment for inflation and (b) when inflation is considered?
Show both hand and spreadsheet solutions.
Solution by Hand
(a) Without infl ation adjustment: The semiannual dividend is I [(50,000)(0.10)]2 $2500.
At a nominal 4% per 6 months for 30 periods,
PW 2500(PA,4%,30) 50,000(PF,4%,30) $58,645
(b) With infl ation: Use the inflated rate i f .
i f 0.04 0.025 (0.04)(0.025) 0.066 per semiannual period
PW 2500(PA,6.6%,30) 50,000(PF,6.6%,30)
2500(12.9244) 50,000(0.1470)
$39,660
Solution by Spreadsheet
Both (a) and (b) require simple, single-cell functions on a spreadsheet (Figure 14–2). Without
an inflation adjustment, the PV function is developed at the nominal 4% rate for 30 periods;
with inflation considered the rate is i f 6.6%, as determined above.
Comment
The $18,985 difference in PW values illustrates the tremendous negative impact made by only
2.5% inflation each 6 months (5.06% per year). Purchasing the $50,000 bond means receiving
$75,000 in dividends over 15 years and the $50,000 principal in year 15. Yet, this is worth only
$39,660 in constant-value dollars.
Figure 14–2
PW computation of a bond purchase (a) without and (b) with an inflation adjustment,
Example 14.2.

14.2 Present Worth Calculations Adjusted for Inflation 373
EXAMPLE 14.3
A self-employed chemical engineer is on contract with Dow Chemical, currently working in a
relatively high-inflation country in Central America. She wishes to calculate a project’s PW
with estimated costs of $35,000 now and $7000 per year for 5 years beginning 1 year from now
with increases of 12% per year thereafter for the next 8 years. Use a real interest rate of 15%
per year to make the calculations (a) without an adjustment for inflation and (b) considering
inflation at a rate of 11% per year.
Solution
(a) Figure 14–3 presents the cash flows. The PW without an adjustment for inflation is found
using i 15% and g 12% in Equations [2.34] and [2.35] for the geometric series.
PW 35,000 7000(PA,15%,4)
{ 7000 [ 1 (
—— 1.12
1.15 ) 9 ]
———————— } (PF,15%,4)
0.15 0.12
35,000 19,985 28,247
$83,232
In the PA factor, n 4 because the $7000 cost in year 5 is the A 1 term in Equation [2.34].
PW = ?
PW g = ?
i =15% per year
0 1 2 3 4 5 6 7 8 9 10 11 12
0 1 2 3 4 5 6 7 8
13
9
Year
Geometric series year
$7000
$7840
$35,000
12% increase
per year
$17,331
Figure 14–3
Cash flow diagram, Example 14.3.
(b) To adjust for inflation, calculate the inflated interest rate by Equation [14.6] and use it to
calculate PW.
i f 0.15 0.11 (0.15)(0.11) 0.2765
PW 35,000 7000(PA,27.65%,4)
{ 7000 [ 1 (
——— 1.12
1.2765 ) 9 ]
————————— } (PF,27.65%,4)
0.2765 0.12
−35,000 7000(2.2545) 30,945(0.3766)
$62,436
Comment
This result demonstrates that in a high-inflation economy, when negotiating the amount of the
payments to repay a loan, it is economically advantageous for the borrower to use future
(inflated) dollars whenever possible to make the payments. The present value of future inflated
dollars is significantly less when the inflation adjustment is included. And the higher the inflation
rate, the larger the discounting because the PF and PA factors decrease in size.

374 Chapter 14 Effects of Inflation
Examples 14.2 and 14.3 above add credence to the “buy now, pay later” philosophy. However, at
some point, the debt-ridden company or individual will have to pay off the debts and the accrued
interest with the inflated dollars. If cash is not readily available at that time, the debts cannot be repaid.
This can happen, for example, when a company unsuccessfully launches a new product, when there
is a serious downturn in the economy, or when an individual loses a salary. In the longer term, this
buy now, pay later approach must be tempered with sound financial practices now, and in the future.
14.3 Future Worth Calculations
Adjusted for Inflation
In future worth calculations, a future amount F can have any one of four different interpretations:
Case 1. The actual amount of money that will be accumulated at time n .
Case 2. The purchasing power of the actual amount accumulated at time n , but stated in
today’s (constant-value) dollars.
Case 3. The number of future dollars required at time n to maintain the same purchasing
power as today; that is, inflation is considered, but interest is not.
Case 4. The amount of money required at time n to maintain purchasing power and earn a
stated real interest rate.
Depending upon which interpretation is intended, the F value is calculated differently, as
described below. Each case is illustrated.
Case 1: Actual Amount Accumulated It should be clear that F, the actual amount of money
accumulated, is obtained using the inflation-adjusted (market) interest rate.
F P (1 i f ) n P( F P,i f ,n) [14.7]
For example, when we quote a market rate of 10%, the inflation rate is included. Over a 7-year
period, $1000 invested at 10% per year will accumulate to
F 1000( F P ,10%,7) $1948
Case 2: Constant-Value Dollars with Purchasing Power The purchasing power of future
dollars is determined by first using the market rate i f to calculate F and then deflating the
future amount through division by (1 f )
n
.
F P(1 i f
————
)n
(1 f) n P (FP, i f ,n)
——————
(1 f ) n [14.8]
This relation, in effect, recognizes the fact that inflated prices mean $1 in the future purchases
less than $1 now. The percentage loss in purchasing power is a measure of how much less. As an
illustration, consider the same $1000 now, and a 10% per year market rate, which includes an
inflation rate of 4% per year. In 7 years, the purchasing power has risen, but only to $1481.
1000 (FP, 10%, 7)
F ————————
(1.04) $1948
7 ———
1.3159 $1481
This is $467 (or 24%) less than the $1948 actually accumulated at 10% (case 1). Therefore, we
conclude that 4% inflation over 7 years reduces the purchasing power of money by 24%.
Also for case 2, the future amount of money accumulated with today’s buying power could equivalently
be determined by calculating the real interest rate and using it in the F P factor to compensate
for the decreased purchasing power. This real interest rate is the i in Equation [14.6].
i f i f if
i (1 f ) f
i i f f ———
1 f
[14.9]

14.3 Future Worth Calculations Adjusted for Inflation 375
The real interest rate i represents the rate at which today’s dollars expand with their same purchasing
power into equivalent future dollars. An inflation rate larger than the market interest
rate leads to a negative real interest rate. The use of this interest rate is appropriate for calculating
the future worth of an investment (such as a savings account or money market fund) when
the effect of inflation must be removed. For the example of $1000 in today’s dollars from
Equation [14.9]
i ————— 0.10 0.04
0.0577 (5.77%)
1 0.04
F 1000( F P ,5.77%,7) $1481
The market interest rate of 10% per year has been reduced to a real rate that is less than 6% per
year because of the erosive effects of 4% per year inflation.
Case 3: Future Amount Required, No Interest This case recognizes that prices increase
when inflation is present. Simply put, future dollars are worth less, so more are needed. No interest
rate is considered in this case—only inflation. This is the situation if someone asks, “How
much will a car cost in 5 years if its current cost is $20,000 and its price will increase by the inflation
rate of 6% per year?” (The answer is $26,765.) No interest rate—only inflation—is involved.
To find the future cost, substitute f for the interest rate in the F P factor.
F P (1 f ) n P ( F P, f,n) [14.10]
Reconsider the $1000 used previously. If it is escalating at exactly the inflation rate of 4% per
year, the amount 7 years from now will be
F 1000( F P ,4%,7) $1316
Case 4: Inflation and Real Interest This is the case applied when a market MARR is established.
Maintaining purchasing power and earning interest must account for both increasing
prices (case 3) and the time value of money. If the growth of capital is to keep up, funds must
grow at a rate equal to or above the real interest rate i plus the inflation rate f . Thus, to make a real
rate of return of 5.77% when the inflation rate is 4%, i f is the market (inflation-adjusted) rate that
must be used. For the same $1000 amount,
i f 0.0577 0.04 0.0577(0.04) 0.10
F 1000( F P ,10%,7) $1948
This calculation shows that $1948 seven years in the future will be equivalent to $1000 now with
a real return of i 5.77% per year and inflation of f 4% per year.
Table 14–3 summarizes which rate is used in the equivalence formulas for the different interpretations
of F . The calculations made in this section explain the following:
• The amount of $1000 now at a market rate of 10% per year will accumulate to $1948 in 7 years.
• The $1948 will have the purchasing power of $1481 of today’s dollars if f 4% per year.
• An item with a cost of $1000 now will cost $1316 in 7 years at an inflation rate of 4% per year.
• It will take $1948 of future dollars to be equivalent to $1000 now at a real interest rate of 5.77%
with inflation considered at 4% per year.
Most corporations evaluate alternatives at a MARR large enough to cover inflation plus some
return greater than their cost of capital, and significantly higher than the safe investment return of
approximately 3.5% mentioned earlier. Therefore, for case 4, the resulting MARR will normally
be higher than the market rate i f . Define the symbol MARR f as the inflation-adjusted or market
MARR, which is calculated in a fashion similar to i f .
MARR f i f i ( f ) [14.11]

376 Chapter 14 Effects of Inflation
TABLE 14–3
Calculation Methods for Various Future Worth Interpretations
Future Worth
Desired
Case 1: Actual dollars
accumulated
Case 2: Purchasing power
of accumulated dollars in
terms of constant-value
dollars
Case 3: Dollars required for
same purchasing power
Case 4: Future dollars to
maintain purchasing
power and to earn a return
Method of
Calculation
Use stated market
rate i f in equivalence
formulas
Use market rate i f in
equivalence and
divide by (1 f ) n
or
Use real i
Use f in place of i in
equivalence
formulas
Calculate i f and use in
equivalence
formulas
Example for
P $1000, n 7,
i f 10%, f 4%
F 1000(FP,10%,7)
1000 (FP,10%,7)
F ————————
(1.04) 7
or
F 1000(FP,5.77%,7)
F 1000(FP,4%,7)
F 1000(FP,10%,7)
The real rate of return i used here is the required rate for the corporation relative to its cost of
capital. Now the future worth F , or FW, is calculated as
F P (1 MARR f ) n P ( F P ,MARR f , n )
For example, if a company has a WACC (weighted average cost of capital) of 10% per year and
requires that a project return 3% per year above its WACC, the real return is i 13%. The
inflation-adjusted MARR is calculated by including the inflation rate of, say, 4% per year. Then
the project PW, AW, or FW will be determined at the rate obtained from Equation [14.11].
MARR f 0.13 0.04 0.13(0.04) 0.1752 (17.52%)
EXAMPLE 14.4
Abbott Mining Systems wants to determine whether it should upgrade a piece of equipment
used in deep mining operations in one of its international operations now or later. If the company
selects plan A, the upgrade will be purchased now for $200,000. However, if the company
selects plan I, the purchase will be deferred for 3 years when the cost is expected to rise
to $300,000. Abbott is ambitious; it expects a real MARR of 12% per year. The inflation rate
in the country has averaged 3% per year. From only an economic perspective, determine
whether the company should purchase now or later (a) when inflation is not considered and
(b) when inflation is considered.
Solution
(a) Infl ation not considered: The real rate, or MARR, is i 12% per year. The cost of plan I is
$300,000 three years hence. Calculate the FW value for plan A three years from now and
select the lower cost.
FW A 200,000(FP,12%,3) $280,986
FW I $−300,000
Select plan A (purchase now).
(b) Infl ation considered: This is case 4; the real rate (12%), and inflation of 3% must be
accounted for. First, compute the inflation-adjusted MARR by Equation [14.11].
MARR f 0.12 0.03 0.12(0.03) 0.1536

14.4 Capital Recovery Calculations Adjusted for Inflation 377
Use MARR f to compute the FW value for plan A in future dollars.
FW A 200,000(FP,15.36%,3) $307,040
FW I $−300,000
Purchase later (plan I) is now selected, because it requires fewer equivalent future dollars.
The inflation rate of 3% per year has raised the equivalent future worth of costs by 9.3%
from $280,986 to $307,040. This is the same as an increase of 3% per year, compounded
over 3 years, or (1.03) 3 1 9.3%.
Most countries have inflation rates in the range of 2% to 8% per year, but hyperinflation is a
problem in countries where political instability, overspending by the government, weak international
trade balances, etc., are present. Hyperinflation rates may be very high—10% to 100% per
month . In these cases, the government may take drastic actions: redefine the currency in terms of
the currency of another country, control banks and corporations, and control the flow of capital
into and out of the country in order to decrease inflation.
In a hyperinflated environment, people usually spend all their money immediately since the
cost of goods and services will be so much higher the next month, week, or day. To appreciate
the disastrous effect of hyperinflation on a company’s ability to keep up, we can rework Example
14.4 b using an inflation rate of 10% per month, that is, a nominal 120% per year (not
considering the compounding effect of inflation). The FW A amount skyrockets and plan I is a
clear choice. Of course, in such an environment the $300,000 purchase price for plan I three
years hence would obviously not be guaranteed, so the entire economic analysis is unreliable.
Good economic decisions in a hyperinflated economy are very difficult to make using traditional
engineering economy methods, since the estimated future values are totally unreliable
and the future availability of capital is uncertain.
14.4 Capital Recovery Calculations
Adjusted for Inflation
It is particularly important in capital recovery (CR) calculations used for AW analysis to include
inflation because current capital dollars must be recovered with future inflated dollars. Since future
dollars have less buying power than today’s dollars, it is obvious that more dollars will be
required to recover the present investment. This suggests the use of the inflated interest rate in the
A P formula. For example, if $1000 is invested today at a real interest rate of 10% per year when
the inflation rate is 8% per year, the equivalent amount that must be recovered each year for
5 years in future dollars is
A 1000( A P ,18.8%,5) $325.59
On the other hand, the decreased value of dollars through time means that investors can spend
fewer present (higher-value) dollars to accumulate a specified amount of future (inflated) dollars.
This suggests the use of a higher interest rate, that is, the i f rate, to produce a lower A value in the
A F formula. The annual equivalent (with adjustment for inflation) of F $1000 five years
from now in future dollars is
A 1000( A F ,18.8%,5) $137.59
For comparison, the equivalent annual amount to accumulate F $1000 at a real i
10% (without adjustment for inflation) is 1000( A F ,10%,5) $163.80. When F is a future
known cost, uniformly distributed payments should be spread over as long a time period as
possible so that the leveraging effect of inflation will reduce the effective annual payment
($137.59 versus $163.80 here).

378 Chapter 14 Effects of Inflation
EXAMPLE 14.5
What annual deposit is required for 5 years to accumulate an amount of money with the same
purchasing power as $680.58 today, if the market interest rate is 10% per year and inflation is
8% per year?
Solution
First, find the actual number of inflated dollars required 5 years in the future that is equivalent
to $680.58 today. This is case 3; Equation [14.10] applies.
F (present purchasing power)(1 f ) 5 680.58(1.08) 5 $1000
The actual amount of the annual deposit is calculated using the market interest rate of 10%.
This is case 4 where A is calculated for a given F.
A 1000(AF,10%,5) $163.80
Comment
The real interest rate is i 1.85% as determined using Equation [14.9]. To put these calculations
into perspective, if the inflation rate is zero when the real interest rate is 1.85%, the future
amount of money with the same purchasing power as $680.58 today is obviously $680.58.
Then the annual amount required to accumulate this future amount in 5 years is A
680.58(AF,1.85%,5) $131.17. This is $32.63 lower than the $163.80 calculated above for
f 8%. This difference is due to the fact that during inflationary periods, dollars deposited now
have more purchasing power than the dollars returned at the end of the period. To make up the
purchasing power difference, more higher-value dollars are required. That is, to maintain
equivalent purchasing power at f 8% per year, an extra $32.63 per year is required.
The logic discussed here explains why, in times of increasing inflation, lenders of money
(credit card companies, mortgage companies, and banks) tend to further increase their market
interest rates. People tend to pay off less of their incurred debt at each payment because they
use any excess money to purchase additional items before the price is further inflated. Also, the
lending institutions must have more money in the future to cover the expected higher costs of
lending money. All this is due to the spiraling effect of increasing inflation. Breaking this cycle
is difficult to do at the individual level and much more difficult to alter at a national level.
CHAPTER SUMMARY
Inflation, treated computationally as an interest rate, makes the cost of the same product or service
increase over time due to the decreased value of money. There are several ways to consider
inflation in engineering economy computations in terms of today’s (constant-value) dollars and
in terms of future dollars. Some important relations are the following:
Inflated interest rate: i f i f if
Real interest rate: i ( i f − f ) (1 f )
PW of a future amount with inflation considered: P F ( P F , i f , n )
Future worth in constant-value dollars of a present amount with the same purchasing power:
F P ( F P , i , n )
Future amount to cover a current amount with inflation only: F P ( F P , f , n )
Future amount to cover a current amount with inflation and interest: F P ( F P , i f , n )
Annual equivalent of a future amount: A F ( A F , i f , n )
Annual equivalent of a present amount in future dollars: A P ( A P , i f , n )
Hyperinflation implies very high f values. Available funds are expended immediately because
costs increase so rapidly that larger cash inflows cannot offset the fact that the currency is losing
value. This can, and usually does, cause a national financial disaster when it continues over extended
periods of time.

Problems 379
PROBLEMS
Interest Rate and Currency Considerations with
Inflation
14.1 What is the difference between today’s dollars and
constant-value dollars ( a ) when using today as the
reference point in time and ( b ) when using 2 years
ago as the reference point?
14.2 State the conditions under which the market interest
rate is ( a ) higher than, ( b ) lower than, and
( c ) the same as the real interest rate.
14.3 What annual inflation rate is implied from an
inflation-adjusted interest rate of 10% per year,
when the real interest rate is 4% per year?
14.4 Determine the inflation-adjusted interest rate for a
growth company that wants to earn a real rate of
return of 20% per year when the inflation rate is
5% per year.
14.5 For a high-growth company that wants to make a
real rate of return of 30% per year, compounded
monthly, determine the inflation-adjusted nominal
interest rate per year. Assume the inflation rate is
1.5% per month.
14.6 A high-tech company whose stock trades on the
NASDAQ stock exchange uses a MARR of 35% per
year. If the chief financial officer (CFO) said the
company expects to make a real rate of return of
25% per year on its investments over the next 3-year
period, what is the company expecting the inflation
rate per year to be over that time period?
14.7 Calculate the inflation-adjusted interest rate per
quarter when the real interest rate is 4% per quarter
and the inflation rate is 1% per quarter.
14.8 Calculate the real interest rate per month if the
nominal inflation-adjusted interest rate per year,
compounded monthly, is 18% and the inflation rate
per month is 0.5%.
14.9 A southwestern city has a contract with Firestone
Vehicle Fleet Maintenance to provide maintenance
services on its fleet of city-owned vehicles such as
street sweepers, garbage trucks, backhoes, and
carpool vehicles. The contract price is fixed at
$45,000 per year for 4 years. If you were asked to
convert the future amounts into constant-value
amounts per today’s dollars, what would the respective
amounts be? Assume the current market
interest rate of 10% per year and inflation rate of
5% per year are expected to remain the same over
the next 4-year period.
14.10 When the inflation rate is 5% per year, how many
inflated dollars will be required 10 years from now
to buy the same things that $10,000 buys now?
14.11 Assume that you want to retire 30 years from now
with an amount of money that will have the same
value (same purchasing power) as $1.5 million
today. If you estimate the inflation rate will be 4%
per year, how many future (then-current) dollars
will you need?
Adjusting for Inflation
14.12 If the inflation rate is 7% per year, how many years
will it take for the cost of something to double when
prices increase at exactly the same rate as inflation?
14.13 The inflation rate in a Central American country is
6% per year. What real rate of return will an investor
make on a $100,000 investment in a copper
mine stock that yields an overall internal rate of
return of 28% per year?
14.14 During periods of hyperinflation, prices increase
rapidly over short periods of time. In 1993, the
Consumer Price Index (CPI) in Brazil was
113.6 billion. In 1994, the CPI was 2472.4 billion.
(a)
What was the inflation rate per year between
1993 and 1994?
( b) Assuming the inflation rate calculated in
part ( a ) occurred uniformly throughout the
year and represented a nominal rate, what
were the monthly and daily inflation rates
over that time period?
14.15 A trust was set up by your grandfather that states
you are to receive $250,000 exactly 5 years from
today. Determine the buying power of the $250,000
in terms of today’s dollars if the market interest
rate is 10% per year and the inflation rate is 4% per
year.
14.16 Assume the inflation rate is 4% per year and the
market interest rate is 5% above the inflation rate.
Determine ( a ) the number of constant-value dollars
5 years in the future that is equivalent to
$30,000 now and ( b ) the number of future dollars
that will be equivalent to $30,000 now.
14.17 The Pell Grant program of the federal government
provides financial aid to needy college students. If
the average grant is slated to increase from $4050
to $5400 over the next 5 years “to keep up with
inflation,” what is the average inflation rate per
year expected to be?

380 Chapter 14 Effects of Inflation
14.18 Ford Motor Company announced that the price of
its F-150 pickup trucks is going to increase by only
the inflation rate for the next 3 years. If the current
price of a well-equipped truck is $28,000 and the
inflation rate averages 2.1% per year, what is the
expected price of a comparably equipped truck
next year? 3 years from now?
14.19 In an effort to reduce pipe breakage, water hammer,
and product agitation, a chemical company plans to
install several chemically resistant pulsation dampeners.
The cost of the dampeners today is $120,000,
but the company has to wait until a permit is approved
for its bidirectional port-to-plant product
pipeline. The permit approval process will take at
least 2 years because of the time required for preparation
of an environmental impact statement.
Because of intense foreign competition, the manufacturer
plans to increase the price only by the inflation
rate each year. If the inflation rate is 2.8% per
year and the company’s MARR is 20% per year,
estimate the cost of the dampeners in 2 years in
terms of ( a ) today’s dollars and ( b ) future dollars.
14.20 A machine currently under consideration by
Holzmann Industries has a cost of $45,000. When
the purchasing manager complained that a similar
machine the company purchased 5 years ago was
much cheaper, the salesman responded that the
cost of the machine has increased solely in accordance
with the inflation rate, which has averaged
3% per year. When the purchasing manager
checked the invoice for the machine he purchased
5 years ago, he saw that the price was $29,000.
Was the salesman telling the truth about the increase
in the cost of the machine? If not, what
should the machine cost now, provided the price
increased by only the inflation rate?
14.21 A report by the National Center for Public Policy
and Higher Education stated that tuition and fees
(T&F) at public colleges and universities increased
by 439% over the last 25 years. During this same
time period, the median family income (MFI) rose
147%. When the report was written, T&F at a
4-year public university constituted 28% of the
MFI of $52,000 (tuition and fees at a private university
constituted 76% of MFI).
(a) What was the tuition and fee cost per year
when the report was written?
(b) What was the T&F cost 25 years ago?
(c) What percentage was T&F of the MFI
25 years ago?
14.22 The headline on a Chronicle of Higher Education
article reads “College Costs Rise Faster than Inflation.”
The article states that tuition at public colleges
and universities increased by 58% over the
past 5 years. ( a ) What was the average annual percentage
increase over that period of time? (b) If the
real increase in tuition (i.e., without inflation) was
5% per year, what was the inflation rate per year?
14.23 Stadium Capital Financing Group helps cashstrapped
sports programs through the marketing of
its “sports mortgage.” At the University of Kansas,
Jayhawks fans can sign up to pay $105,000 over
10 years for the right to buy top seats for football
during the next 30 years. In return, the season tickets
will stay locked in at current-year prices. Season
tickets in tier 1 are currently selling for $350.
A fan plans to purchase the sports mortgage along
with season tickets now and each year for the next
30 years (31 seasons). What is the dollar amount of
the savings on the tickets (with no interest considered),
if ticket prices rise at a rate of 3% per year
for the next 30 years?
Present Worth Calculations with Inflation
14.24 There are two ways to account for inflation in present
worth calculations. What are they?
14.25 An environmental testing company needs to purchase
equipment 2 years from now and expects to
pay $50,000 at that time . At a real interest rate of
10% per year and inflation rate of 4% per year, what
is the present worth of the cost of the equipment?
14.26 Carlsbad Gas and Electric is planning to purchase
a degassing tower for removing CO 2 from acidified
saltwater. The supplier quoted a price of
$125,000 if the unit is purchased within the next
3 years. Your supervisor has asked you to calculate
the present worth of the tower when considering
inflation. Assuming the tower will not be purchased
for 3 years, calculate the present worth at
an interest rate of 10% per year and an inflation
rate of 4% per year.
14.27 How much can the manufacturer of superconducting
magnetic energy storage systems afford to
spend now on new equipment in lieu of spending
$75,000 four years from now? The company’s real
MARR is 12% per year, and the inflation rate is
3% per year.
14.28 Find the present worth of the cash flows shown.
Some are expressed as constant-value (CV) dollars,
and others are inflated dollars. Assume a real
interest rate of 8% per year and an inflation rate of
6% per year.
Year 1 2 3 4 5
Cash Flow, $ 3000 6000 8000 4000 5000
Stated as CV Inflated Inflated CV CV

Problems 381
14.29 A doctor is on contract to a medium-sized oil company
to provide medical services at remotely located,
widely separated refineries. The doctor is
considering the purchase of a private plane to reduce
the total travel time between refineries. The doctor
can buy a used Lear jet now or wait for a new very
light jet (VLJ) that will be available 3 years from
now. The cost of the VLJ will be $1.9 million, payable
when the plane is delivered in 3 years. The doctor
has asked you, his friend, to determine the present
worth of the VLJ so that he can decide whether to
buy the used Lear now or wait for the VLJ. If the
MARR is 15% per year and the inflation rate is projected
to be 3% per year, what is the present worth of
the VLJ with inflation considered?
14.30 A regional infrastructure building and maintenance
contractor is trying to decide whether to buy
a new compact horizontal directional drilling
(HDD) machine now or wait to buy it 2 years from
now (when a large pipeline contract will require
the new equipment). The HDD machine will include
an innovative pipe loader design and
maneuverable undercarriage system. The cost of
the system is $68,000 if purchased now or $81,000
if purchased 2 years from now. At a real MARR of
10% per year and an inflation rate of 5% per year,
determine if the company should buy now or later
( a ) without any adjustment for inflation and
( b ) with inflation considered.
14.31 An engineer must recommend one of two rapidprototyping
machines for integration into an upgraded
manufacturing line. She obtained estimates
from salespeople from two companies. Salesman
A gave her the estimates in constant-value
(today’s) dollars, while saleswoman B provided
the estimates in future (then-current) dollars. The
company’s MARR is equal to the real rate of return
of 20% per year, and inflation is estimated at
4% per year. Use PW analysis to determine which
machine the engineer should recommend.
A
(in CV Dollars)
B
(in Future Dollars)
First cost, $ 140,000 155,000
AOC, $ per year 25,000 40,000
Life, years 10 10
14.32 A salesman from Industrial Water Services (IWS),
who is trying to get his foot in the door for a large
account in Fremont, offered water chlorination
equipment for $2.1 million. This is $400,000 more
than the price offered by a competing saleswoman
from AG Enterprises. However, IWS said the company
would not have to pay for the equipment until
the warranty runs out. If the equipment has
a 2-year warranty, determine which offer is better.
The company’s real MARR is 12% per year, and
the inflation rate is 4% per year.
14.33 A chemical engineer is considering two sizes of
pipes, small (S) and large (L), for moving distillate
from a refinery to the tank farm. A small pipeline
will cost less to purchase (including valves and
other appurtenances) but will have a high head loss
and, therefore, a higher pumping cost. In writing
the report, the engineer compared the alternatives
based on future worth values, but the company
president wants the costs expressed as present dollars.
Determine present worth values if future
worth values are FW S $2.3 million and FW L
$2.5 million. The company uses a real interest rate
of 1% per month and an inflation rate of 0.4% per
month. Assume the future worth values were for a
10-year project period.
14.34 As an innovative way to pay for various software
packages, a new high-tech service company has
offered to pay your company, Custom Computer
Services (CCS), in one of three ways: (1) pay
$480,000 now, (2) pay $1.1 million 5 years from
now, or (3) pay an amount of money 5 years from
now that will have the same purchasing power as
$850,000 now. If you, as president of CCS, want to
earn a real interest rate of 10% per year when the
inflation rate is 6% per year, which offer should
you accept?
Future Worth and Other Calculations with Inflation
14.35 If the inflation rate is 6% per year and a person
wants to earn a true (real) interest rate of 10% per
year, determine the number of future dollars she
has to receive 10 years from now if the present investment
is $10,000.
14.36 How many future dollars would you need 5 years
from now just to have the same buying power as
$50,000 now, if the deflation rate is 3% per year?
14.37 If a company deposits $100,000 into an account
that earns a market interest rate of 10% per year at
a time when the defl ation rate is 1% per year, what
will be the purchasing power of the accumulated
amount (with respect to today’s dollars) at the end
of 15 years?
14.38 Harmony Corporation plans to set aside $60,000
per year beginning 1 year from now for replacing
equipment 5 years from now. What will be the purchasing
power (in terms of current-value dollars)
of the amount accumulated, if the investment
grows by 10% per year, but inflation averages
4% per year?

382 Chapter 14 Effects of Inflation
14.39 The strategic plan of a solar energy company that
manufactures high-efficiency solar cells includes
an expansion of its physical plant in 4 years. The
engineer in charge of planning estimates the expenditure
required now to be $8 million, but in
4 years, the cost will be higher by an amount equal
to the inflation rate. If the company sets aside
$7,000,000 now into an account that earns interest
at 7% per year, what will the inflation rate have to
be in order for the company to have exactly the
right amount of money for the expansion?
14.40 Five years ago, an industrial engineer deposited
$10,000 into an account and left it undisturbed
until now. The account is now worth $25,000.
(a) What was the overall ROR during the
5 years?
(b) If the inflation over that period was 4% per
year, what was the real ROR?
(c) What is the buying power of the $25,000 now
compared to the buying power 5 years ago?
14.41 A Toyota Tundra can be purchased today for
$32,350. A civil engineering firm is going to need
three more trucks in 2 years because of a land development
contract it just won. If the price of the
truck increases exactly in accordance with an estimated
inflation rate of 3.5% per year, determine
how much the three trucks will cost in 2 years.
14.42 The Nobel Prize is administered by The Nobel
Foundation, a private institution that was founded
in 1900 based on the will of Alfred Nobel, the inventor
of dynamite. In part, his will stated: “The
capital shall be invested by my executors in safe
securities and shall constitute a fund, the interest
on which shall be annually distributed in the form
of prizes to those who, during the preceding year,
shall have conferred the greatest benefit on mankind.”
The will further stated that the prizes were
to be awarded in physics, chemistry, peace, physiology
or medicine, and literature. In addition to a
gold medal and a diploma, each recipient receives
a substantial sum of money that depends on the
Foundation’s income that year. The first Nobel
Prize was awarded in 1901 in the amount of
$150,000. In 1996, the award was $653,000; it was
$1.4 million in 2009.
(a) If the increase between 1996 and 2009 was
strictly due to inflation, what was the average
inflation rate per year during that 13-year
period?
(b) If the Foundation expects to invest money
with a return of 5% above the inflation rate,
how much will a laureate receive in 2020,
provided the inflation rate averages 3% per
year between 2009 and 2020?
14.43 Timken Roller Bearing is a manufacturer of
seamless tubes for drill bit collars. The company
is planning to add larger-capacity robotic arms to
one of its assembly lines 3 years from now. If it
is done now, the cost of the equipment with installation
is $2.4 million. If the company’s real
MARR is 15% per year, determine the equivalent
amount the company can spend 3 years from
now in then-current dollars if the inflation rate is
2.8% per year.
14.44 The data below show two patterns of inflation that
are exactly the opposite of each other over a
20-year time period.
(a) If each machine costs $10,000 in year 0 and
they both increase in cost exactly in accordance
with the inflation rate, how much will
each machine cost at the end of year 20?
(b) What is the average inflation rate over the
time period for machine A (that is, what single
inflation rate would result in the same
final cost for machine A)?
(c) In which years will machine A cost more
than machine B?
Year Machine A, % Machine B, %
1 10 2
2 10 2
3 2 10
4 2 10
5 10 2
6 10 2
7 2 10
8 2 10
19 2 10
20 2 10
14.45 Factors that increase costs and prices—especially
for materials and manufacturing costs sensitive to
market, technology, and labor availability—can be
considered separately using the real interest rate i ,
the inflation rate f , and additional increases that
grow at a geometric rate g . The future amount is
calculated based on a current estimate by using the
relation
F P (1 i ) n (1 f ) n (1 g ) n
P [(1 i ) (1 f ) (1 g )]
n
The product of the first two terms enclosed in parentheses
results in the inflated interest rate i f . The
geometric rate is the same one used in the geometric
series (Chapter 2). It commonly applies to
maintenance and repair cost increases as machinery
ages. This is over and above the inflation rate.
If the current cost to manufacture an electronic
subcomponent is $300,000 per year, what is the

Problems 383
equivalent cost in 3 years, provided the average
annual rates for i , f , and g are 10%, 3%, and 2%,
respectively?
Capital Recovery with Inflation
14.46 An electric utility is considering two alternatives
for satisfying state regulations regarding pollution
control for one of its generating stations.
This particular station is located at the outskirts
of a major U.S. city and a short distance from a
large city in a neighboring country. The station is
currently producing excess VOCs and oxides of
nitrogen. Two plans have been proposed for satisfying
the regulators. Plan A involves replacing
the burners and switching from fuel oil to natural
gas. The cost of the option will be $300,000 initially
and an extra $900,000 per year in fuel
costs. Plan B involves going to the foreign city
and running gas lines to many of the “backyard”
brick-making sites that now use wood, tires, and
other combustible waste materials for firing the
bricks. The idea behind plan B is that by reducing
the particulate pollution responsible for smog
in the neighboring city, there would be greater
benefit to U.S. citizens than would be achieved
through plan A. The initial cost of plan B will be
$1.2 million for installation of the lines. Additionally,
the electric company would subsidize
the cost of gas for the brick makers to the extent
of $200,000 per year. Extra air monitoring associated
with this plan will cost an additional
$150,000 per year. For a 10-year project period
and no salvage value for either plan, which one
should be selected on the basis of an annual
worth analysis at a real interest rate of 7% per
year and an inflation rate of 4% per year?
14.47 A chemical engineer who is smitten with wanderlust
wants to build a reserve fund that will be sufficient
to permit her to take time off from work to
travel around the world. Her goal is to save enough
money over the next 3 years so that when she begins
her trip, the amount she has accumulated will
have the same buying power as $72,000 today. If
she expects to earn a market rate of 12% per year
on her investments and inflation averages 5% per
year over the 3 years, how much must she save
each year to reach her goal?
14.48 An entrepreneur engaged in wildcat oil well drilling
is seeking investors who will put up $500,000
for an opportunity to reap high returns if the venture
is successful. The prospectus states that a
return of at least 22% per year for 5 years is
likely. How much will the investors have to
receive each year to recover their money if an
inflation rate of 5% per year is to be included in
the calculation?
14.49 Veri-Trol, Inc. manufactures in situ calibration
verification systems that confirm flow measurement
accuracies without removing the meters. The
company has decided to modify the main assembly
line with one of the enhancement processes
shown below. If the company’s real MARR is 15%
per year, which process has the lower annual cost
when inflation of 5% per year is considered?
Process X
Process Y
First cost, $ 65,000 90,000
Operating cost, $ per year 40,000 34,000
Salvage value, $ 0 10,000
Life, years 5 5
14.50 Johnson Thermal Products used austenitic
nickel-chromium alloys to manufacture resistance
heating wire. The company is considering a new
annealing-drawing process to reduce costs. If the
new process will cost $3.7 million now, how much
must be saved each year to recover the investment
in 5 years if the company’s MARR is a real 12%
per year and the inflation rate is 3% per year?
14.51 The costs associated with a small X-ray inspection
system are $40,000 now and $24,000 per year,
with a $6000 salvage value after 3 years. Determine
the equivalent cost of the system if the real
interest rate is 10% per year and the inflation rate
is 4% per year.
14.52 Maintenance costs for pollution control equipment
on a pulverized coal cyclone furnace are expected
to be $180,000 now and another $70,000 three
years from now. The CFO of Monongahela Power
wants to know the equivalent annual cost of the
equipment in years 1 through 5. If the company
uses a real interest rate of 9% per year and the inflation
rate averages 3% per year, what is the
equivalent annual cost of the equipment?
14.53 MetroKlean LLC, a hazardous waste soil cleaning
company, borrowed $2.5 million for 5 years to finance
start-up costs for a new project involving
site reclamation. The company expects to earn a
real rate of return of 20% per year. The average
inflation rate is 5% per year.
(a)
Determine the capital recovery required each
year with inflation considered.
(b) Determine the capital recovery if the company
is satisfied with accumulating $2.5 million
at the end of the 5 years with inflation
considered.
(c) Determine the capital recovery in part ( b )
without considering inflation.

384 Chapter 14 Effects of Inflation
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
14.54 Inflation occurs when:
(a) Productivity increases
(b) The value of the currency decreases
(c) The value of the currency increases
(d) The price of gold decreases
14.55 “Then-current” (future) dollars can be converted
into constant-value dollars by:
(a) Multiplying them by (1 f ) n
(b) Multiplying them by (1 f ) n n
(1 i )
(c) Dividing them by (1 f ) n
(d)
n
Dividing them by (1 i f )
14.56 To calculate how much something will cost if you
expect its cost to increase by exactly the inflation
rate, you should:
(a) Multiply by (1 f ) n
(b) Multiply by [(1 f ) n n
(1 i ) ]
(c) Divide by (1 f ) n
(d) Multiply by (1 i f ) n
14.57 If the market interest rate is 16% per year when the
inflation rate is 9% per year, the real interest rate is
closest to:
( a ) 6.4%
( b ) 7%
( c ) 9%
( d ) 15.6%
14.58 The market interest rate i f is 12% per year, compounded
semiannually. For an inflation rate of 2%
per 6 months, the effective semiannual interest rate
is closest to:
( a ) 2%
( b ) 3%
( c ) 4%
( d ) 6%
14.59 Construction equipment has a cost today of
$40,000. If its cost has increased only by the inflation
rate of 6% per year, when the market interest
rate was 10% per year, its cost 10 years ago was
closest to:
(a) $15,420
(b) $22,335
(c) $27,405
(d) $71,630
14.60 The amount of money that would be accumulated
now from an investment of $1000 25 years ago at
a market rate of 5% per year and an inflation rate
averaging 2% per year over that time period is
closest to:
(a) $1640
( b) $3385
( c) $5430
( d) Over $5500
14.61 If $1000 is invested now, the number of future dollars
required 10 years from now to earn a real interest
rate of 6% per year, when the inflation rate is
4% per year, is closest to:
(a) $1480
(b) $1790
(c) $2650
(d) Over $2700
14.62 If you expect to receive a gift of $50,000 six years
from now, the present worth of the gift at a real
interest rate of 4% per year and an inflation rate of
3% per year is closest to:
(a) $27,600
(b) $29,800
(c) $33,100
(d) $37,200
14.63 New state-mandated emission testing equipment
for annual inspection of automobiles at Charlie’s
Garage has a first cost of $30,000, an annual operating
cost of $7000, and a $5000 salvage value
after its 20-year life. For a real interest rate of 5%
per year and an inflation rate of 4% per year, the
annual capital recovery requirement for the equipment
(in future dollars) is determined by:
(a) AW 30,000( A P ,4%,20) 7000
5000( A F ,4%,20)
(b) AW 30,000( A P ,5%,20) 7000
5000( A F ,5%,20)
(c) AW 30,000( A P ,9%,20) 7000
5000( A F ,9%,20)
(d) AW 30,000( A P ,9.2%,20) 7000
5000( A F ,9.2%,20)
14.64 Temporary price deflation may occur in specific
sectors of the economy for all of the following reasons
except:
(a) Improved technology
(b) Excessive demand
(c) Dumping
(d) Increased productivity

Case Study 385
CASE STUDY
INFLATION VERSUS STOCK AND BOND INVESTMENTS
Background
The savings and investments that an individual maintains
should have some balance between equity (corporate stocks
that rely on market growth and dividend income) and fixedincome
investments (bonds that pay dividends to the
purchaser and a guaranteed amount upon maturity). When
inflation is moderately high, bonds offer a low return relative
to stocks, because the potential for market growth is not present
with bonds. Additionally, the forces of inflation make the
dividends worth less in future years, because there is no inflation
adjustment made in the amount the dividend pays as time
passes. However, bonds do offer a steady income that may be
important to an individual, and they serve to preserve the
principal invested in the bond, because the face value is returned
at maturity.
Information
Earl is an engineer who wants a predictable flow of money
for travel and vacations. He has a collection of stocks in his
retirement portfolio, but no bonds. He has accumulated a
total of $50,000 of his own funds in low-yielding savings accounts
and wants to improve his long-term return from this
nonretirement program “nest egg.” He can choose additional
stocks or bonds, but has decided to not split the $50,000 between
the two forms of investments. There are two choices he
has outlined, with the best estimates he can make at this time.
He assumes the effects of federal and state income taxes will
be the same for both forms of investment.
Stock purchase: Stocks purchased through a mutual fund
would pay an estimated 2% per year dividend and appreciate
in value at 5% per year.
Bond purchase: If he purchased a bond, he would have a
predictable income of 5% per year and the $50,000 face
value after the 12-year maturity period.
Case Study Questions
The analysis that Earl has laid out has the following questions.
Can you answer them for him for both choices?
1. What is the overall rate of return after 12 years?
2. If he decided to sell the stock or bond immediately after
the fifth annual dividend, what is his minimum selling
price to realize a 7% real return? Include an adjustment
of 4% per year for inflation.
3. If Earl needed some money in the future, say, immediately
after the fifth dividend payment, what would be
the minimum selling price in future dollars, if he were
only interested in recovering an amount that maintained
the purchasing power of the original price?
4. As a follow-on to question 3, what happens to the
selling price (in future dollars) 5 years after purchase,
if Earl is willing to remove (net out) the future purchasing
power of each of the dividends in the computation
to determine the required selling price 5 years
hence?
5. Earl plans to keep the stocks or bonds for 12 years, that
is, until the bond matures. However, he wants to make
the 7% per year real return and make up for the expected
4% per year inflation. For what amount must he
sell the stocks after 12 years, or buy the bonds now to
ensure he realizes this return? Do these amounts seem
reasonable to you, given your knowledge of the way
that stocks and bonds are bought and sold?

CHAPTER 15
Cost
Estimation and
Indirect Cost
Allocation
LEARNING OUTCOMES
Purpose: Make cost estimates using different methods; demonstrate the allocation of indirect costs using traditional and
activity-based costing rates.
SECTION TOPIC LEARNING OUTCOME
15.1 Approach • Explain the bottom-up and design-to-cost (topdown)
approaches to cost estimation.
15.2 Unit method • Use the unit method to make a preliminary cost
estimate.
15.3 Cost index • Use a cost index to estimate a present cost based
on historical data.
15.4 Cost capacity • Use a cost-capacity equation to estimate
component, system, or plant costs.
15.5 Factor method • Estimate total plant cost using the factor
method.
15.6 Indirect cost rates • Allocate indirect costs using traditional indirect
cost rates.
15.7 ABC allocation • Use the Activity-Based Costing (ABC) method to
allocate indirect costs.
15.8 Ethics and profit • Describe how biased estimation can become an
ethical dilemma.

U
p to this point, cost and revenue cash flow values have been stated or assumed
as known. In reality, they are not; they must be estimated. This chapter explains
what cost estimation involves and applies cost estimation techniques. Cost estimation
is important in all aspects of a project, but especially in the stages of project conception,
preliminary design, detailed design, and economic analysis. When a project is developed
in the private or the public sector, questions about costs and revenues will be posed by
individuals representing many different functions: management, engineering, construction,
production, quality, finance, safety, environmental, legal, and marketing, to name some. In
engineering practice, the estimation of costs receives much more attention than revenue
estimation; costs are the topic of this chapter.
Unlike direct costs for labor and materials, indirect costs are not easily traced to a specific
department, machine, or processing line. Therefore, allocation of indirect costs for functions
such as utilities, safety, management and administration, purchasing, and quality is made
using some rational basis. Both the traditional method of allocation and the Activity-Based
Costing (ABC) method are covered in this chapter.
15.1 Understanding How Cost Estimation
Is Accomplished
Cost estimation is a major activity performed in the initial stages of virtually every effort in
industry, business, and government. In general, most cost estimates are developed for either a
project or a system; however, combinations of these are very common. A project usually involves
physical items, such as a building, bridge, manufacturing plant, or offshore drilling
platform, to name just a few. A system is usually an operational design that involves processes,
services, software, and other nonphysical items. Examples might be a purchase order
system, a software package, an Internet-based remote-control system, or a health care delivery
system. Of course, many projects will have major elements that are not physical, so estimates
of both types must be developed. For example, consider a computer network system.
There would be no operational system if only the costs of computer hardware plus wire and
wireless connectors were estimated; it is equally important to estimate the software, personnel,
and maintenance costs.
Thus far virtually all cash flow estimates in the examples, problems, progressive examples,
and case studies were stated or assumed to be known. In real-world practice, the cash flows for
costs and revenues must be estimated prior to the evaluation of a project or comparison of alternatives.
We concentrate on cost estimation because costs are the primary values estimated for the
economic analysis. Revenue estimates utilized by engineers are usually developed in marketing,
sales, and other departments.
Costs are comprised of direct costs and indirect costs . Normally direct costs are estimated
with some detail, then the indirect costs are added using standard rates and factors.
However, direct costs in many industries, including manufacturing and assembly settings,
have become a small percentage of overall product cost, while indirect costs have become
much larger. Accordingly, many industrial settings require some estimating for indirect costs
as well. Indirect cost allocation is discussed in detail in later sections of this chapter. Primarily,
direct costs are discussed here.
Because cost estimation is a complex activity, the following questions form a structure for the
discussion that follows.
DirectIndirect costs
• What cost components must be estimated?
• What approach to cost estimation will be applied?
• How accurate should the estimates be?
• What estimation techniques will be utilized?
Costs to Estimate If a project revolves around a single piece of equipment, for example, an
industrial robot, the cost components will be significantly simpler and fewer than the components
for a complete system such as the manufacturing and testing line for a new product. Therefore, it
is important to know up front how much the cost estimation task will involve. Examples of cost
components are the first cost P and the annual operating cost (AOC), also called the M&O costs

388 Chapter 15 Cost Estimation and Indirect Cost Allocation
(maintenance and operating) of equipment. Each component will have several cost elements .
Listed below are sample elements of the first cost and AOC components.
First cost component P:
Elements:
Equipment cost
Delivery charges
Installation cost
Insurance coverage
Initial training of personnel for equipment use
Delivered-equipment cost is the sum of the first two elements in the list above; installedequipment
cost adds the third element.
AOC component (part of the equivalent annual cost A):
Elements: Direct labor cost for operating personnel
Direct materials
Maintenance costs (daily, periodic, repairs, etc.)
Rework and rebuild
Some of these elements, such as equipment cost, can be determined with high accuracy; others,
such as maintenance costs, are harder to estimate. When costs for an entire system must be estimated,
the number of cost components and elements is likely to be in the hundreds. It is then
necessary to prioritize the estimation tasks.
For familiar projects (houses, office buildings, highways, and some chemical plants) there are
standard cost estimation software packages available. For example, state highway departments
utilize software that prompts for the cost components (bridges, pavement, cut-and-fill profiles,
etc.) and estimates costs with time-proven, built-in relations. Once these components are estimated,
exceptions for the specific project are added. However, there are no “canned” software
packages for a large percentage of industrial, business, service and public sector projects.
Cost Estimation Approach Historically in industry, business, and the public sector, a
bottom-up approach to cost estimation was applied. For a simple rendition of this approach, see
Figure 15–1 (left side). The progression is as follows: cost components and their elements are
identified, cost elements are estimated, and estimates are summed to obtain total direct cost. The
price is then determined by adding indirect costs and the profit margin, which is usually a percentage
of the total cost.
The bottom-up approach treats the required price as an output variable and the cost estimates
as input variables. This approach works well when competition is not a dominant factor in pricing
the product or service.
Figure 15–1 (right side) shows a simplistic progression for the design-to-cost , or top-down, approach.
The competitive price establishes the target cost.
The design-to-cost, or top-down, approach treats the competitive price as an input variable
and the cost estimates as output variables. This approach is useful in encouraging innovation,
new design, manufacturing process improvement, and efficiency. These are some of the essentials
of value engineering and value-added systems engineering.
This approach places greater emphasis on the accuracy of the price estimation activity. The target
cost must be realistic, or else it can be a disincentive to design and engineering staff. The designto-cost
approach is best applied in the early stages of a new or enhanced product design. The
detailed design and specific equipment options are not yet known, but the price estimates assist
in establishing target costs for different components.
Usually, the resulting approach is some combination of these two philosophies. However, it is
helpful to understand up front what approach is to be emphasized. Historically, the bottom-up
approach was more predominant in Western engineering cultures, especially in North America.
The design-to-cost approach is considered routine in Eastern engineering cultures; however,

15.1 Understanding How Cost Estimation Is Accomplished 389
Required
price
Desired
profit
Top-down approach Figure 15–1
Simplified cost estimation
Allowed
profit
Competitive
price
processes for bottom-up
and top-down approaches.
Total cost
Target cost
Indirect
costs
Maintenance
and operations
Direct
costs
Indirect
costs
Maintenance
and operations
Cost
component
estimates
Direct
labor
Direct
materials
Direct
labor
Direct
materials
Cost
component
estimates
Equipment and
capital recovery
Equipment
capital recovery
At design stage
Bottom-up approach
Before design stage
Design-to-cost approach
globalization of engineering design has speeded the adoption of the design-to-cost approach
worldwide.
Accuracy of the Estimates No cost estimates are expected to be exact; however, they are
expected to be reasonable and accurate enough to support economic scrutiny. The accuracy required
increases as the project progresses from preliminary design to detailed design and on to
economic evaluation. Cost estimates made before and during the preliminary design stage are
expected to be good “first-cut” estimates that serve as input to the project budget.
When utilized at early and conceptual design stages, estimates are referred to as order-ofmagnitude
estimates and generally range within 20% of actual cost. At the detailed design
stage, cost estimates are expected to be accurate enough to support economic evaluation for a
go–no go decision. Every project setting has its own characteristics, but a range of 5% of actual
costs is expected at the detailed design stage. Figure 15–2 shows the general range of estimate
accuracy for the construction cost of a building versus time spent in preparing the estimate. Obviously,
the desire for better accuracy has to be balanced against the cost of obtaining it.
Accuracy of Estimate
30%
25%
20%
15%
10%
5%
}
10 minutes
1 hour
1 day
Scoping/feasibility
Order of magnitude estimate
Partially designed
Design 60–100% complete
Detailed estimate
1 week
Time Spent on Estimate
3 weeks
Figure 15–2
Characteristic curve of
estimate accuracy versus
time spent to estimate
construction cost of a
building.

390 Chapter 15 Cost Estimation and Indirect Cost Allocation
Cost Estimation Techniques Methods such as expert opinion and comparison with comparable
installations serve as excellent estimators. The use of the unit method and cost indexes base
the present estimate on past cost experiences, with inflation considered. Models such as costcapacity
equations and the factor method are simple mathematical techniques applied at the
preliminary design stage. They are called cost estimating relationships (CERs). There are many
additional methods discussed in the handbooks and publications of different industries.
Most cost estimates made in a professional setting are accomplished in part or wholly using
software packages linked to updated databases that contain cost indexes and rates for the locations,
products, or processes being studied. There are a wide variety of estimators, cost trackers,
and cost compliance software systems, most of them developed for specific industries. Corporations
usually standardize on one or two packages to ensure consistency over time and projects.
15.2 Unit Method
The unit method is a popular preliminary estimation technique applicable to virtually all professions.
The total estimated cost C T is obtained by multiplying the number of units N by a per unit
cost factor u .
C T u N [15.1]
Unit cost factors must be updated frequently to remain current with changing costs, areas,
and inflation. Some sample unit cost factors (and values) are
Total average cost of operating an automobile (52¢ per mile)
Cost to bury fiber cable in a suburban area ($30,000 per mile)
Cost to construct a parking space in a parking garage ($4500 per space)
Cost of constructing interstate highway ($6.2 million per mile)
Cost of house construction per livable area ($225 per square foot)
Applications of the unit method to estimate costs are easily found. If house construction costs
average $225 per square foot, a preliminary cost estimate for an 1800-square-foot house, using
Equation [15.1], is $405,000. Similarly, a 200-mile trip should cost about $104 for the car only
at 52¢ per mile.
When there are several components to a project or system, the unit cost factors for each component
are multiplied by the amount of resources needed, and the results are summed to obtain
the total cost C T . This is illustrated in Example 15.1.
EXAMPLE 15.1
Justin, an ME with Dynamic Castings, has been asked to make a preliminary estimate of the
total cost to manufacture 1500 sections of high-pressure gas pipe using an advanced centrifugal
casting method. Since a 20% estimate is acceptable at this preliminary stage, a unit
method estimate is sufficient. Use the following resource and unit cost factor estimates to
help Justin.
Materials: 3000 tons at $45.90 per ton
Machinery and tooling: 1500 hours at $120 per hour
Direct labor in plant:
Casting and treating: 3000 hours at $55 per hour
Finishing and shipping: 1200 hours at $45 per hour
Indirect labor: 400 hours at $75 per hour
Solution
Apply Equation [15.1] to each of the five areas and sum the results to obtain the total cost
estimate of $566,700. Table 15–1 provides the details.

15.3 Cost Indexes 391
TABLE 15–1
Total Cost Estimate Using Unit Cost Factors for Several Resource Areas,
Example 15.1
Resource Amount N Unit Cost Factor u, $ Cost Estimate, u N, $
Materials 3000 tons 45.90 per ton 137,700
Machinery, tooling 1500 hours 120 per hour 180,000
Labor, casting 3000 hours 55 per hour 165,000
Labor, finishing 1200 hours 45 per hour 54,000
Labor, indirect 400 hours 75 per hour 30,000
Total cost estimate 566,700
15.3 Cost Indexes
This section explains indexes and their use in cost estimation. An index is a ratio or other number
based on observation and used as an indicator or measure. A preliminary cost estimate is often
based on a cost index.
A cost index is a ratio of the cost of something today to its cost sometime in the past. The index is
dimensionless and measures relative cost change over time . Because these indexes are sensitive to
technological change, the predefined quantity and quality of elements used to define the index may be
hard to retain over time, thus causing “index creep.” Timely updating of the index is very important.
One such index that most people are familiar with is the Consumer Price Index (CPI), which
shows the relationship between present and past costs for many of the things that “typical” consumers
must buy. This index includes such items as rent, food, transportation, and certain services.
Other indexes track the costs of equipment, and goods and services that are more pertinent
to the engineering disciplines. Table 15–2 is a listing of some of the more common indexes.
TABLE 15–2
Types and Sources of Various Cost Indexes
Type of Index
Overall prices
Consumer (CPI)
Producer (wholesale)
Construction
Chemical plant overall
Equipment, machinery, and supports
Construction labor
Buildings
Engineering and supervision
Engineering News Record overall
Construction
Building
Common labor
Skilled labor
Materials
EPA treatment plant indexes
Large-city advanced treatment (LCAT)
Small-city conventional treatment (SCCT)
Federal highway
Contractor cost
Equipment
Marshall and Swift (M&S) overall
M&S specific industries
Labor
Output per worker-hour by industry
Source
Bureau of Labor Statistics
U.S. Department of Labor
Chemical Engineering
Engineering News Record (ENR)
Environmental Protection Agency
and ENR
Marshall & Swift
U.S. Department of Labor

392 Chapter 15 Cost Estimation and Indirect Cost Allocation
TABLE 15–3
Values for Selected Indexes
Year
CE Plant
Cost Index
ENR Construction
Cost Index
M&S Equipment
Cost Index
1995 381.1 5471 1027.5
1996 381.8 5620 1039.2
1997 386.5 5826 1056.8
1998 389.5 5920 1061.9
1999 390.6 6059 1068.3
2000 394.1 6221 1089.0
2001 394.3 6343 1093.9
2002 395.6 6538 1104.2
2003 401.7 6694 1123.6
2004 444.2 7115 1178.5
2005 468.2 7446 1244.5
2006 499.6 7751 1302.3
2007 525.4 7967 1373.3
2008 575.4 8310 1449.3
2009 521.9 8570 1468.6
2010 (midyear) 555.3 8837 1461.3
The general equation for updating costs through the use of a cost index over a period from
time t 0 (base) to another time t is
EXAMPLE 15.2
where C t estimated cost at present time t
C 0 cost at previous time t 0
I t index value at time t
I 0 index value at time t 0
C t C 0 ( I t
—
I 0
) [15.2]
Generally, the indexes for equipment and materials are made up of a mix of components that
are assigned certain weights, with the components sometimes further subdivided into more basic
items. For example, the equipment, machinery, and support component of the chemical plant cost
index is subdivided into process machinery, pipes, valves and fittings, pumps and compressors,
and so forth. These subcomponents, in turn, are built up from even more basic items such as pressure
pipe, black pipe, and galvanized pipe. Table 15–3 shows the Chemical Engineering plant
cost index (CEPCI), the Engineering News Record (ENR) construction cost index, and the
Marshall and Swift (M&S) equipment cost index for several years. The base period of 1957 to
1959 is assigned a value of 100 for the CEPCI, 1913 100 for the ENR index, and 1926 100
for the M&S equipment cost index.
Current and past values of several of the indexes may be obtained from the Internet (usually
for a fee). For example, the CE plant cost index is available at www.che.compci. The ENR construction
cost index is found at www.construction.com. This latter site offers a comprehensive
series of construction-related resources, including several ENR cost indexes and cost estimation
systems. A website used by many engineering professionals in the form of a “technical chat
room” for all types of topics, including estimation, is www.eng-tips.com.
In evaluating the feasibility of a major construction project, an engineer is interested in estimating
the cost of skilled labor for the job. The engineer finds that a project of similar complexity
and magnitude was completed 5 years ago at a skilled labor cost of $360,000. The ENR
skilled labor index was 3496 then and is now 5127. What is the estimated skilled labor cost for
the new project?

15.3 Cost Indexes 393
Solution
The base time t 0 is 5 years ago. Using Equation [15.2], the present cost estimate is
C r 360,000 (
——— 5127
3496 )
$527,952
In the manufacturing and service industries, tabulated cost indexes are not readily available.
The cost index will vary, perhaps with the region of the country, the type of product or service,
and many other factors. When estimating costs for a manufacturing system, for example, it is
often necessary to develop the cost index for high-priority items such as subcontracted components,
selected materials, and labor costs. The development of the cost index requires the actual
cost at different times for a prescribed quantity and quality of the item. The base period is a selected
time when the index is defined with a basis value of 100 (or 1). The index each year (period)
is determined as the cost divided by the base-year cost and multiplied by 100 (or 1). Future
index values may be forecast using simple extrapolation or more refined mathematical techniques,
such as time-series analysis.
EXAMPLE 15.3
Sean, owner of Alamo Pictures, makes fact-based documentaries about the Old West and
sells them in a variety of outlets, by mail, and online. He has decided to expand into new
areas and wants to make cost estimates for three of the more significant labor costs involved
in making these types of films. The Director of Finance of Alamo Pictures generated the
annual average hourly costs (Table 15–4).
(a) Make 2008 the base year, and determine the cost indexes using a basis of 1.00. Comment
on the trend of each index over the years.
(b) Sean expects to utilize a lot of graphics services in 2014 for a planned documentary; however,
it is the most rapidly increasing cost component. The cost in 2010 was $78 per hour;
assume a worst-case scenario is that the graphics index continues the same arithmetic trend
it had from 2010 to 2011. Determine the hourly cost that he should budget for in 2014.
Solution
(a) For each type of service, calculate I t I 0 where t 2005, 2006, . . . with 2008 as the base
year 0. Table 15–5 presents the indexes. Observations about trend are as follows:
Graphics labor cost: Constantly increasing over all years.
Stuntmen labor cost: Rising until 2009, then stable.
Actors labor cost: Higher in 2007 and 2008; comparatively lower and stable in other years.
TABLE 15–4 Average Hourly Costs for Three Services, Example 15.3
Cost of Service, Average $ per Hour
Type of Service 2005 2006 2007 2008 2009 2010 2011
Graphics 50 56 65 67 70 78 90
Stuntmen 50 55 60 70 87 83 85
Actors 80 80 90 90 80 75 85
TABLE 15–5 Index Values with 2008 as Base Year, Example 15.3
Index I t I 0
Type of Service 2005 2006 2007 2008 2009 2010 2011
Graphics 0.75 0.84 0.97 1.00 1.04 1.16 1.34
Stuntmen 0.71 0.79 0.86 1.00 1.24 1.19 1.21
Actors 0.89 0.89 1.00 1.00 0.89 0.83 0.94

394 Chapter 15 Cost Estimation and Indirect Cost Allocation
(b) The index in 2010 is 1.16. The increase to 2011 is 0.18; the index value in 2014 will be
1.34 3(0.18) 1.88. Equation [15.2] finds the expected, worst-case cost in 2014.
C 2014 C 2010 (I 2014 I 2010 ) 78(1.881.16)
78(1.62)
$126 per hour
15.4 Cost-Estimating Relationships:
Cost-Capacity Equations
Design variables (speed, weight, thrust, physical size, etc.) for plants, equipment, and construction
are determined in the early design stages. Cost-estimating relationships (CERs) use these
design variables to predict costs. Thus, a CER is generically different from the cost index method,
because the index is based on the cost history of a defined quantity and quality of a variable.
One of the most widely used CER models is a cost-capacity equation . As the name implies,
an equation relates the cost of a component, system, or plant to its capacity. This is also known
as the power law and sizing model . Since many cost-capacity equations plot as a straight line on
log-log paper, a common form is
C 2 C 1 ( Q x
2
——
Q 1
)
[15.3]
where C 1 cost at capacity Q 1
C 2 cost at capacity Q 2
x correlating exponent
The value of the exponent for various components, systems, or entire plants can be obtained or
derived from a number of sources, including Plant Design and Economics for Chemical Engineers,
Preliminary Plant Design in Chemical Engineering, Chemical Engineers’ Handbook,
technical journals (especially Chemical Engineering ), the U.S. Environmental Protection Agency,
professional or trade organizations, consulting firms, handbooks, and equipment companies.
Table 15–6 is a partial listing of typical values of the exponent for various units. When an exponent
value for a particular unit is not known, it is common practice to use the value of x 0.6. In
fact, in the chemical processing industry, Equation [15.3] is referred to as the six-tenths model.
The exponent x in the cost-capacity equation is commonly in the range 0 x 1.
If x 1, economies of scale provide a cost advantage for larger sizes.
If x 1, a linear relationship is present.
If x 1, there are diseconomies of scale present in that a larger size is more costly than that
of a linear relation.
It is especially powerful to combine the time adjustment of the cost index ( I t I 0 ) from Equation
[15.2] with a cost-capacity equation to estimate costs that change over time. If the index is
embedded into the cost-capacity computation in Equation [15.3], the cost at time t and capacity
level 2 may be written as the product of two independent terms.
C 2, t (cost at time 0 of level 2) (time adjustment cost index)
[
C 1,0 ( Q x
2
——
Q 1
) I 0
)
] ( I t
—
This is commonly expressed without the time subscripts. Thus,
C 2 C 1 ( Q x
2
——
Q 1
)
The following example illustrates the use of this relation.
( I t
—
I 0
) [15.4]

15.5 Cost-Estimating Relationships: Factor Method 395
TABLE 15–6
Sample Exponent Values for Cost-Capacity Equations
ComponentSystemPlant Size Range Exponent
Activated sludge plant 1–100 MGD 0.84
Aerobic digester 0.2–40 MGD 0.14
Blower 1000–7000 ftmin 0.46
Centrifuge 40–60 in 0.71
Chlorine plant 3000–350,000 tonsyear 0.44
Clarifier 0.1–100 MGD 0.98
Compressor, reciprocating (air service) 5–300 hp 0.90
Compressor 200–2100 hp 0.32
Cyclone separator 20–8000 ft 3 min 0.64
Dryer 15–400 ft 2 0.71
Filter, sand 0.5–200 MGD 0.82
Heat exchanger 500–3000 ft 2 0.55
Hydrogen plant 500–20,000 scfd 0.56
Laboratory 0.05–50 MGD 1.02
Lagoon, aerated 0.05–20 MGD 1.13
Pump, centrifugal 10–200 hp 0.69
Reactor 50–4000 gal 0.74
Sludge drying beds 0.04–5 MGD 1.35
Stabilization pond 0.01–0.2 MGD 0.14
Tank, stainless 100–2000 gal 0.67
Note: MGD million gallons per day; hp horsepower; scfd standard cubic feet per day.
EXAMPLE 15.4
The total design and construction cost for a digester to handle a flow rate of 0.5 million gallons
per day (MGD) was $1.7 million in 2010. Estimate the cost today for a flow rate of 2.0 MGD.
The exponent from Table 15–6 for the MGD range of 0.2 to 40 is 0.14. The cost index in 2010
of 131 has been updated to 225 for this year.
Solution
Equation [15.3] can estimate the cost of the larger system in 2010, but it must be updated by
the cost index to today’s dollars. Equation [15.4] performs both operations at once. The estimated
cost is
C 2 1,700,000 (
—— 2.0
0.5 ) 0.14 (
—— 225
131 )
1,700,000(1.214)(1.718) $3.546 million
15.5 Cost-Estimating Relationships:
Factor Method
Another widely used model for preliminary cost estimates of process plants is called the factor
method . While the methods discussed above can be used to estimate the costs of major items of
equipment, processes, and the total plant costs, the factor method was developed specifically for
total plant costs . The method is based on the premise that fairly reliable total plant costs can be
obtained by multiplying the cost of the major equipment by certain factors. Since major equipment
costs are readily available, rapid plant estimates are possible if the appropriate factors are
known. These factors are commonly referred to as Lang factors after Hans J. Lang, who first
proposed the method.

396 Chapter 15 Cost Estimation and Indirect Cost Allocation
In its simplest form, the factor method is expressed in the same form as the unit method
C T hC E [15.5]
DirectIndirect costs
where C T total plant cost
h overall cost factor or sum of individual cost factors
C E total cost of major equipment
The h may be one overall cost factor or, more realistically, the sum of individual cost components
such as construction, maintenance, direct labor, materials, and indirect cost elements. This follows
the cost estimation approaches presented in Figure 15–1.
In his original work, Lang showed that direct cost factors and indirect cost factors can be
combined into one overall factor for some types of plants as follows: solid process plants, 3.10;
solid-fluid process plants, 3.63; and fluid process plants, 4.74. These factors reveal that the total
installed-plant cost is many times the first cost of the major equipment.
EXAMPLE 15.5
An engineer with Valero Petroleum has learned that an expansion of the solid-fluid process
plant is expected to have a delivered equipment cost of $2.08 million. If the overall cost factor
for this type of plant is 3.63, estimate the plant’s total cost.
Solution
The total plant cost is estimated by Equation [15.5].
C T 3.63(2,080,000)
$7,550,400
Subsequent refinements of the factor method have led to the development of separate factors
for direct and indirect cost components. Direct costs as discussed in Section 15.1 are specifically
identifiable with a product, function, or process. Indirect costs are not directly attributable to a
single function, but are shared by several because they are necessary to perform the overall objective.
Examples of indirect costs are general administration, computer services, quality, safety,
taxes, security, and a variety of support functions. The factors for both direct and indirect costs
are sometimes developed for use with delivered-equipment costs and other times for installedequipment
costs, as defined in Section 15.1. In this text, we assume that all factors apply to delivered-equipment
costs, unless otherwise specified.
For indirect costs, some of the factors apply to equipment costs only, while others apply to the
total direct cost. In the former case, the simplest procedure is to add the direct and indirect cost
factors before multiplying by the delivered-equipment cost. The overall cost factor h can be
written as
n
h 1 f i [15.6]
where f i factor for each cost component
i 1 to n components, including indirect cost
If the indirect cost factor is applied to the total direct cost, only the direct cost factors are added
to obtain h . Therefore, Equation [15.5] is rewritten as
C T
[
C E (
where f I indirect cost factor
f i factors for direct cost components
Examples 15.6 and 15.7 illustrate these equations.
i1
n
1 f i ) ] (1 f I ) [15.7]
i1

15.6 Traditional Indirect Cost Rates and Allocation 397
EXAMPLE 15.6
The delivered-equipment cost for a small chemical process plant is expected to be $2 million.
If the direct cost factor is 1.61 and the indirect cost factor is 0.25, determine the total plant cost.
Solution
Since all factors apply to the delivered-equipment cost, they are added to obtain h, the total cost
factor in Equation [15.6].
The total plant cost is
h 1 1.61 0.25 2.86
C T 2.86(2,000,000) $5,720,000
EXAMPLE 15.7
A new container-handling crane at the Port of Singapore is expected to have a deliveredequipment
cost of $875,000. The cost factor for the installation of tracks, concrete, steel, noise
abatement, supports, etc., is 0.49. The construction factor is 0.53, and the indirect cost factor is
0.21. Determine the total cost if (a) all cost factors are applied to the cost of the delivered
equipment and (b) the indirect cost factor is applied to the total direct cost.
Solution
(a) Total equipment cost is $875,000. Since both the direct and indirect cost factors are applied
to only the equipment cost, the overall cost factor from Equation [15.6] is
The total cost is
h 1 0.49 0.53 0.21 2.23
C T 2.23(875,000) $1,951,250
(b) Now the total direct cost is calculated first, and Equation [15.7] is used to estimate total
cost.
n
h 1 f i 1 0.49 0.53 2.02
i1
C T [875,000(2.02)](1.21) $2,138,675
Comment
Note the decrease in estimated cost when the indirect cost is applied to the equipment cost only
in part (a). This illustrates the importance of determining exactly what the factors apply to
before they are used.
15.6 Traditional Indirect Cost Rates and Allocation
Costs incurred in the production of an item or delivery of a service are tracked and assigned by a
cost accounting system . For the manufacturing environment, it can be stated generally that the
statement of cost of goods sold (discussed in Appendix B) is one end product of this system. The
cost accounting system accumulates material costs, labor costs, and indirect costs (also called
overhead costs or factory expenses) by using cost centers. All costs incurred in one department
or process line are collected under a cost center title, for example, Department 3X. Since direct
materials and direct labor are usually directly assignable to a cost center, the system need only
identify and track these costs. Of course, this in itself is no easy chore, and the cost of the tracking
system may prohibit collection of all direct cost data to the level of detail desired.
One of the primary and more difficult tasks of cost accounting is the allocation of indirect costs
when it is necessary to allocate them separately to departments, processes, and product lines.

398 Chapter 15 Cost Estimation and Indirect Cost Allocation
TABLE 15–7
Sample Indirect Cost Allocation Bases
Cost Category
Taxes
Heat, light
Power
Receiving, purchasing
Personnel, machine shop
Building maintenance
Software
Quality control
Possible Allocation Basis
Space occupied
Space, usage, number of outlets
Space, direct labor hours, horsepower-hours, machine hours
Cost of materials, number of orders, number of items
Direct labor hours, direct labor cost
Space occupied, direct labor cost
Number of accesses
Number of inspections
Indirect costs
Indirect costs are costs associated with property taxes, service and maintenance departments,
personnel, legal, quality, supervision, purchasing, utilities, software development,
etc. They must all be allocated to the using cost center. Detailed collection of these data is
cost- prohibitive and often impossible; thus, allocation schemes are utilized to distribute the
expenses on a reasonable basis.
A listing of possible allocation bases is included in Table 15–7. Historically, common bases have
been direct labor cost, direct labor hours, machine-hours, number of employees, space, and direct
materials.
Most allocation is accomplished utilizing a predetermined indirect cost rate, computed by
using the general relation.
Indirect cost rate —————————————
estimated total indirect costs
estimated basis level
[15.8]
EXAMPLE 15.8
The estimated indirect cost is the amount allocated to a cost center. For example, if a division has
two producing departments, the total indirect cost allocated to a department is used as the numerator
in Equation [15.8] to determine the department rate. Example 15.8 illustrates allocation
when the cost center is a machine.
The manager of beauty products at BestWay wants to determine allocation rates for $150,000
of indirect costs for the three machines used to process beauty lotions. The following information
was obtained from last year’s budget for the three machines. Determine rates for each
machine if the amount is equally distributed.
Cost Source Allocation Basis Estimated Activity Level
Machine 1 Direct labor cost $100,000
Machine 2 Direct labor hours 2000 hours
Machine 3 Direct material cost $250,000
Solution
Applying Equation [15.8] for each machine, the annual rates are
indirect budget
Machine 1 rate ———————
direct labor cost ————
50,000
100,000
$0.50 per direct labor dollar
indirect budget
Machine 2 rate ————————
direct labor hours ———
50,000
2000
$25 per direct labor hour
indirect budget
Machine 3 rate ——————— ————
50,000
material cost 250,000
$0.20 per direct material dollar

15.6 Traditional Indirect Cost Rates and Allocation 399
Now the actual direct labor costs and hours and material costs are determined for this
year, and each dollar of direct labor cost spent on machine 1 implies that $0.50 in indirect
cost will be added to the cost of the product. Similarly, indirect costs are added for machines
2 and 3.
When the same allocation basis is used to distribute indirect costs to several cost centers, a
blanket rate may be determined. For example, if direct materials are the basis for allocation to
four separate processing lines, the blanket rate is
Indirect cost rate ———————————
total indirect costs
total direct materials cost
If $500,000 in indirect costs and $3 million in materials are estimated for the four lines, the blanket
indirect rate is 500,0003,000,000 $0.167 per materials cost dollar. Blanket rates are easier
to calculate and apply, but they do not account for differences in the type of activities accomplished
in each cost center.
In most cases, machinery or processes add value to the end product at different rates per unit
or hour of use. For example, light machinery may contribute less per hour than heavy, more expensive
machinery. This is especially true when advanced technology processing, for example,
an automated manufacturing cell, is used along with traditional methods, for example, nonautomated
finishing equipment. The use of blanket rates in these cases is not recommended, as the
indirect cost will be incorrectly allocated. The lower-value-contribution machinery will accumulate
too much of the indirect cost. The approach should be the application of different bases for
different machines, activities, etc., as discussed earlier and illustrated in Example 15.8. The use
of different, appropriate bases is often called the productive hour rate method since the cost
rate is determined based on the value added, not a uniform or blanket rate. Realization that more
than one basis should be normally used in allocating indirect costs has led to the use of activitybased
costing methods, as discussed in the next section.
Once a period of time (month, quarter, or year) has passed, the indirect cost rates are applied
to determine the indirect cost charge, which is then added to direct costs. This results in the total
cost of production, which is called the cost of goods sold, or factory cost . These costs are all accumulated
by cost center .
If the total indirect cost budget is correct, the indirect costs charged to all cost centers for the
period of time should equal this budget amount. However, since some error in budgeting always
exists, there will be overallocation or underallocation relative to actual charges, which is termed
allocation variance . Experience in indirect cost estimation assists in reducing the variance at the
end of the accounting period.
EXAMPLE 15.9
Since the manager determined indirect cost rates for BestWay (Example 15.8), she can now
compute the total factory cost for a month. Perform the computations using the data in
Table 15–8. Also calculate the variance for indirect cost allocation for the month.
Solution
Start with the cost of goods sold (factory cost) relation given by Equation [B.1] in Appendix B,
which is
Cost of goods sold direct materials direct labor indirect costs
TABLE 15–8
Actual Monthly Data Used for Indirect Cost Allocation
Cost Source Machine Number Actual Cost, $ Actual Hours
Material 1 3,800
3 19,550
Labor 1 2,500 650
2 3,200 750
3 2,800 720

400 Chapter 15 Cost Estimation and Indirect Cost Allocation
To determine indirect cost, the rates from Example 15.8 are applied:
Machine 1 indirect (labor cost)(rate) 2500(0.50) $1250
Machine 2 indirect (labor hours)(rate) 750(25.00) $18,750
Machine 3 indirect (material cost)(rate) 19,550(0.20) $3910
Total charged indirect cost $23,910
Factory cost is the sum of actual material and labor costs from Table 15–8 of $31,850 and the
indirect cost charge for a total of $55,760.
Based on the annual indirect cost budget of $150,000, one month represents 112 of the
total or
Monthly budget ————
150,000
12
$12,500
The allocation variance for total indirect cost is
Variance 12,500 23,910 $11,410
This is a large budget underallocation, since much more was actually charged than allocated.
The $12,500 budgeted for the three machines represents a 91.3% underallocation of indirect
costs. This analysis for only one month of a year will most likely prompt a rapid review of the
rates and the indirect cost budget.
EXAMPLE 15.10
Once estimates of indirect costs are determined, it is possible to perform an economic analysis of
the present operation versus a proposed operation. Such a study is described in Example 15.10.
For several years, Cuisinart Corporation has purchased the carafe assembly of its major coffeemaker
line at an annual cost of $2.2 million. The suggestion to make the component in-house
has been made. For the three departments involved, the annual indirect cost rates, estimated
material, labor, and hours are found in Table 15–9. The allocated hours column is the time
necessary to produce the carafes for a year.
Equipment must be purchased with the following estimates: first cost of $2 million, salvage
value of $50,000, and life of 10 years. Perform an economic analysis for the make alternative,
assuming that a market rate of 15% per year is the MARR.
Solution
For making the components in-house, the AOC is comprised of direct labor, direct material,
and indirect costs. Use the data of Table 15–9 to calculate the indirect cost allocation.
Department
Department A: 25,000(10) $250,000
Department B: 25,000(5) 125,000
Department C: 10,000(15) 150,000
$525,000
AOC 500,000 300,000 525,000 $1,325,000
TABLE 15–9 Production Cost Estimates for Example 15.10
Basis,
Hours
Rate
per Hour, $
Indirect Costs
Allocated
Hours
Material
Cost, $
Direct
Labor
Cost, $
A Labor 10 25,000 200,000 200,000
B Machine 5 25,000 50,000 200,000
C Labor 15 10,000 50,000 100,000
300,000 500,000

15.7 Activity-Based Costing (ABC) for Indirect Costs 401
The make alternative annual worth is
AW make P(AP,i,n) S(AF,i,n) AOC
2,000,000(AP,15%,10) 50,000(AF,15%,10) 1,325,000
$1,721,037
Currently, the carafes are purchased with an AW of
AW buy $2,200,000
It is cheaper to make, because the AW of costs is less.
15.7 Activity-Based Costing (ABC) for Indirect Costs
As automation, software, and manufacturing technologies have advanced, the number of
direct labor hours necessary to manufacture a product has decreased substantially. Where
once as much as 35% to 45% of the final product cost was represented in labor, now the
labor component is commonly 5% to 15% of total manufacturing cost. However, the indirect
cost may represent as much as 35% to 45% of the total manufacturing cost. The use of
bases, such as direct labor hours, to allocate indirect cost is not accurate enough for automated
and technologically advanced environments. This has led to the development of
methods that replace or supplement traditional cost allocations that rely upon one form or
another of Equation [15.8]. Also, allocation bases different from traditional ones are commonly
utilized.
A product that by traditional methods may have contributed a large portion to profit may actually
be a loser when indirect costs are allocated more correctly. Companies that have a wide variety
of products and produce some in small lots may find that traditional allocation methods have
a tendency to underallocate the indirect cost to small-lot products. This may indicate that they are
profitable, when in actuality they are losing money.
The best allocation method for high-overhead industries is Activity-Based Costing (ABC). It
is designed to identify cost centers, activities, and cost drivers. Descriptions of each follow.
Cost centers: The final products or services of the corporation are called cost centers or cost
pools. They receive the allocated indirect costs.
Activities: These are usually support departments (purchasing, quality, IT, maintenance, engineering,
supervision) that generate the indirect costs which are then distributed to the cost
centers.
Cost drivers: Commonly expressed in volumes, these drive the consumption of a shared resource.
Examples are the number of purchase orders, cost of engineering change orders,
number of machine setups, number of safety violations, and the like.
Implementing ABC involves several steps.
1. Identify each activity and its total cost.
2. Identify the cost drivers and their usage volumes.
3. Calculate the indirect cost rate for each activity.
ABC indirect cost rate
total cost of activity
————————————
total volume of cost driver
[15.9]
4. Use the rate to allocate indirect cost to cost centers for each activity.
As an illustration, assume a company that produces two types of industrial lasers (cost centers)
has three primary support departments (activities; step 1 in the procedure). The allocation
for costs generated by the purchasing department, for example, is based on the number of purchase
orders (step 2) to support laser production. The ABC rate (step 3) in dollars per purchase
order is used to allocate indirect costs to the two laser products (step 4).

402 Chapter 15 Cost Estimation and Indirect Cost Allocation
EXAMPLE 15.11
A multinational aerospace firm uses traditional methods to allocate manufacturing and management
support costs for its European division. However, accounts such as business travel
have historically been allocated on the basis of the number of employees at the plants in France,
Italy, Germany, and Spain.
The president recently stated that some product lines are likely generating much more travel
than others. The ABC system is chosen to augment the traditional method to more precisely
allocate travel costs to major product lines at each plant.
(a) First, assume that allocation of total observed travel expenses of $500,000 to the plants
using a traditional basis of workforce size is sufficient. If total employment of 29,100 is
distributed as follows, allocate the $500,000.
Paris, France plant
12,500 employees
Florence, Italy plant 8,600 employees
Hamburg, Germany plant 4,200 employees
Barcelona, Spain plant 3,800 employees
(b) Now, assume that corporate management wants to know more about travel expenses based
on product line, not merely plant location and workforce size. The ABC method will be
applied to allocate travel costs to major product lines. Annual plant support budgets indicate
that the following percentages are expended for travel:
Paris 5% of $2 million
Florence 15% of $500,000
Hamburg 17.5% of $1 million
Barcelona 30% of $500,000
Further, the study indicates that in 1 year a total of 500 travel vouchers were processed by
the management of the major five product lines produced at the four plants. The distribution
is as follows:
Paris Product lines—1 and 2; number of vouchers—50 for line 1, 25 for 2.
Florence Product lines—1, 3, and 5; vouchers—80 for line 1, 30 for 3, 30 for 5.
Hamburg Product lines—1, 2 and 4; vouchers—100 for line 1, 25 for 2, 20 for 4.
Barcelona Product line—5; vouchers—140 for line 5.
Use the ABC method to determine how the product lines drive travel costs at the plants.
Solution
(a) In this case, Equation [15.8] takes the form of a blanket rate per employee.
travel budget
Indirect cost rate ———————
total workforce
————
$500,000 $17.1821 per employee
29,100
Using this traditional basis of rate times workforce size results in a plant-by-plant allocation.
Paris $17.1821(12,500) $214,777
Florence $147,766
Hamburg $72,165
Barcelona $65,292
(b) The ABC method is more involved to apply, and the by-plant amounts will be different
from those in part (a) since completely different bases are applied. Use the 4-step procedure
to allocate travel costs to the five products.
Step 1. The total amount to be allocated is determined from the percentages of each
plant’s support budget devoted to travel. The number is determined from the
percent-of-budget data as follows:
0.05(2,000,000) . . . 0.30(500,000) $500,000

15.8 Making Estimates and Maintaining Ethical Practices 403
TABLE 15–10 ABC Allocation of Travel Cost ($ in Thousands), Example 15.11
Product Line
1 2 3 4 5 Total
Paris 50 25 75
Florence 80 30 30 140
Hamburg 100 25 20 145
Barcelona 140 140
Total $230 $50 $30 $20 $170 $500
Step 2.
The cost driver is the number of travel vouchers submitted by the management
unit responsible for each product line at each plant. The allocation will be to
the products directly, not to the plants. However, the travel allocation to the
plants can be determined afterward since we know what product lines are
produced at each plant.
Step 3. Equation [15.9] determines an ABC allocation rate.
ABC allocation rate ———————————
total cost of travel
total number of vouchers
————
$500,000
500
$1000 per voucher
Step 4.
Table 15–10 summarizes the vouchers and allocation by product line and by city.
Product 1 ($230,000) and product 5 ($170,000) drive the travel costs based on
the ABC analysis. Comparison of the by-plant totals in Table 15–10 (far right
column) with the respective totals in part (a) indicates a substantial difference in
the amounts allocated, especially to Paris, Hamburg, and Barcelona. This comparison
verifies the president’s statement that product lines, not plants, drive
travel requirements.
Comment
Let’s assume that product 1 has been produced in small lots at the Hamburg plant for a number
of years. This analysis, when compared to the traditional cost allocation method in part (a), reveals
a very interesting fact. In the ABC analysis, Hamburg has a total of $145,000 travel dollars
allocated, $100,000 from product 1. In the traditional analysis based on workforce size, Hamburg
was allocated only $72,165—about 50% of the more precise ABC analysis amount. This indicates
to management the need to examine the manufacturing lot size practices at Hamburg and
possibly other plants, especially when a product is currently manufactured at more than one plant.
Some proponents of the ABC method recommend discarding the traditional cost accounting
methods of a company and utilizing ABC exclusively. This is not a good approach, since ABC is
not a complete cost system. The ABC method provides information that assists in cost control,
while the traditional method emphasizes cost allocation and cost estimation. The two systems
work well together with the traditional methods allocating costs that have identifiable direct
bases, for example, direct labor. ABC analysis is usually more expensive and time-consuming
than a traditional cost allocation system, but in many cases it can assist in understanding the
economic impact of management decisions and in controlling certain types of indirect costs.
15.8 Making Estimates and Maintaining
Ethical Practices
Making estimates about the future of costs, revenues, cash flows, rates of return, and many other
parameters is routine when one is engaged in any type of economic analysis. Public agencies,
private corporations, and not-for-profit businesses all make economic decisions based on these

404 Chapter 15 Cost Estimation and Indirect Cost Allocation
estimates, most of which are made by employees of the organizations or by outside consultants
hired to perform specific activities under contract. The opportunities for bias, poor accuracy,
deception, and profit-driven or other motives are always present. The personal morals and adherence
to codes of professional ethics discussed in Section 1.3 guide individuals in their work to
make fair and believable estimates for the analyses and decisions that follow.
The NSPE Code of Ethics for Engineers (Appendix C) referenced previously starts with a list
of six Fundamental Canons. One very relevant to estimation integrity is “Avoid deceptive acts.”
Acts that bias the results from experimental samples, previous cost data, or survey results for the
purposes of personal gain, increased profits, or favoritism are examples of unethical behavior.
Estimates of all types should be founded on practices such as the following:
Base estimates on sound information gathered over a range of situations representative of the
current one.
Use accepted theory and techniques in taking statistical samples, building budget elements,
and drawing conclusions that are included in proposals, applications, and recommendations.
As a consultant or contractor, keep personal and working relationships separate when making
estimates and delivering the final documentation to a client or sponsor.
The second case study at the end of this chapter presents an example of some ethical challenges
present when preparing estimates and proposals for contract work.
CHAPTER SUMMARY
Cost estimates are not expected to be exact, but they should be accurate enough to support a
thorough economic analysis using an engineering economy approach. There are bottom-up and
top-down approaches; each treats price and cost estimates differently.
Costs can be updated via a cost index, which is a ratio of costs for the same item at two separate
times. The Consumer Price Index (CPI) is an often-quoted example of cost indexing. Cost
estimating may also be accomplished with a variety of models called cost-estimating relationships.
Two of them are
Cost-capacity equation— good for estimating costs from design variables for equipment, materials,
and construction
Factor method— good for estimating total plant cost
Traditional cost allocation uses an indirect cost rate determined for a machine, department, product
line, etc. Bases such as direct labor cost, direct material cost, and direct labor hours are used.
With increased automation and information technology, different techniques of indirect cost allocation
have been developed. The Activity-Based Costing method is an excellent technique to
augment the traditional allocation method.
The ABC method allocates indirect costs on the rationale that purchase orders, inspections,
machine setups, reworks, etc. drive the costs accumulated in departments or functions, such as
quality, purchasing, accounting, and maintenance. Improved understanding of how the company
or plant accumulates indirect costs is a major by-product of implementing the ABC method.
PROBLEMS
Understanding Cost Estimation
15.1 Rank the following estimate types in terms of time
spent to carry out the estimate (most time to least
time): partially designed, design 60% to 100%
complete, order of magnitude, scopingfeasibility,
detailed estimate.
15.2 Classify the following cost elements as first cost
(FC) components or annual operating cost (AOC)
components for a piece of equipment on the
shop floor: supplies, insurance, equipment cost,
utility cost, installation, delivery charges, labor
cost.
15.3 State whether actual (A) or estimated (E) costs are
more likely to be used to carry out the following
activities: calculate taxes, make bids, pay bonuses,
determine profit or loss, predict sales, set prices,

Problems 405
evaluate proposals, distribute resources, plan production,
and set goals.
15.4 Identify the output and input variables in both the
bottom-up and top-down approaches to cost estimating.
15.5 Classify the following costs as typically direct (D)
or indirect (I):
Project staff
Utilities
Raw materials
Project supplies
Administrative staff
Audit and legal
Rent
Equipment training
Labor
Miscellaneous office supplies
15.6 Identify each of the following costs associated
with owning an automobile as direct or indirect
(assume a direct cost is one that is directly attributable
to the number of miles driven): license plate,
driver’s license, gasoline, highway toll fee, oil
change, repairs after collision, gasoline tax,
monthly loan payment, annual inspection fee, and
garage rental.
15.7 In the early and conceptual design stages of a project,
what are the cost estimates called? Approximately
how close should they be to the actual cost?
Unit Costs
15.8 Use the unit cost method to determine the preliminary
cost of a guardrail for a small bridge if a total
of 120 linear feet will be required at a cost of
$58.19 per linear foot for material, equipment, and
labor.
15.9 Sun-Metro Bus is planning to construct a 600-car
parking garage on the outskirts of the city to encourage
people to use the “Park ’n Ride” to work
or for shopping downtown. Prepare a preliminary
cost estimate for the garage if the cost per parking
space is $4700.
15.10 The Office of the Undersecretary of Defense periodically
releases unit cost data for use in military
construction programs. If the unit cost for a satellite
communications center (with shielding and
generator) is $496 per square foot, what would be
the estimated cost of a 6000-square-foot building?
15.11 The Department of Defense uses area cost factors
(ACFs) to adjust for differences in construction costs
in different parts of the country (and world). The
area cost factor for Andros Island in the Bahamas is
1.70 while the ACF for Rapid City, South Dakota, is
0.93. If a cold storage processing warehouse costs
$1,350,000 in Rapid City, estimate the cost for
Andros Island.
15.12 Preliminary cost estimates for jails can be made
using costs based on either unit area (square feet)
or unit volume (cubic feet). If the unit area cost is
$185 per square foot and the average height of the
ceilings is 10 feet, what is the unit volume cost?
15.13 The unit area and unit volume total project costs
for a library are $114 per square foot and $7.55 per
cubic foot, respectively. Based on these numbers,
what is the average height of library rooms?
15.14 The equipment required to place 160 cubic yards of
concrete by experienced workers is two gasoline engine
vibrators and one concrete pump. If the vibrators
cost $76 per day and the concrete pump costs
$580 per day, estimate ( a ) the cost of the equipment
per cubic yard of concrete and ( b ) the equipment
cost for placing 56 cubic yards of concrete.
15.15 A labor crew for placing concrete consists of
1 labor foreman at $25.85 per hour, 1 cement finisher
at $28.60 per hour, 5 laborers at $23.25 per
hour each, and 1 equipment operator at $31.45 per
hour. Such a crew, called a C20 crew, can place 160
cubic yards of concrete per 8-hour day. Determine
( a ) the cost per day of labor for the C20 crew, ( b ) the
cost of the C20 crew per cubic yard of concrete, and
( c ) the cost to place 250 cubic yards of concrete.
15.16 Site work activities associated with constructing a
small bridge are shown in the table below. The
table includes the quantity of each activity, the unit
of measurement associated with each activity, and
the unit cost of each activity. Use the data to determine
( a ) the total cost for structural excavation, ( b )
the total cost for the pile-driving rig, and ( c ) the
total labor cost for the site work.
Activity
Excavation,
unclassified
Excavation,
structural
Backfill,
compacted
Quantity
Unit of
Measurement
Labor
Unit Cost, $
Equipment
Unit Cost, $
Material
Unit Cost, $
1667 cy 1.35 1.43 0
120 cy 21.31 5.00 0
340 cy 7.78 1.72 0
Pile-driving rig job ls 5688 6420 300
Piling, steel, 2240 lf 3.13 2.93 16.57
driving
Legend: cy cubic yard; ls lump sum; lf linear foot
Cost Indexes
15.17 From historical data, you discover that the national
average construction cost of middle schools is
$10,500 per student. If the state index for Texas is
76.9 and for California it is 108.5, estimate the
total construction cost of a middle school for 800
students in each state.

406 Chapter 15 Cost Estimation and Indirect Cost Allocation
15.18 A consulting engineering firm is preparing a preliminary
cost estimate for a design-construct project
of a coal processing plant. The firm, which
completed a similar project in 2001 with a construction
cost of $30 million, wants to use the ENR
construction cost index to update the cost. Use the
values in Table 15–3 to estimate the cost of construction
for a similar-size plant in mid-2010.
15.19 If the editors at ENR decide to redo the construction
cost index so that the year 2000 has a base
value of 100, determine the value for the year ( a )
1995 and ( b ) 2009.
15.20 (a) Estimate the value of the ENR construction
cost index by using the average (compounded)
percentage change in its value between 1995
and 2005 to predict the value in 2009.
(b) How much difference (numerical) is there between
the estimated and actual 2009 values?
Is this an underestimate or overestimate?
15.21 An engineer who owns a construction company
that specializes in large commercial projects noticed
that material costs increased at a rate of 1%
per month over the past 12 months. If a material
cost index were created for each month of this year
with the value of the index set at 100 for the beginning
of the year, what would be the value of the
index at the end of the year? Express your answer
to two decimal places.
15.22 Using 1913 as the base year with a value of 100,
the ENR construction cost index (CCI) for August
2009 was 8563.35. For August 2010, the CCI
value was 8837.38. ( a ) What was the inflation rate
for construction for that 1-year period? ( b ) What
will be the CCI value in August 2013, provided the
inflation rate remains the same as it was in the
2009–2010 period?
15.23 Electropneumatic general-purpose pressure transducers
convert supply pressure to regulated output
pressure in direct proportion to an electrical input
signal. The cost of a certain transducer was $194 in
1985 when the M&S equipment index was at
789.6. If the price increased exactly in proportion
to the M&S equipment index, what would its cost
be in 2010 (midyear) when the index value was
1461.3?
15.24 The ENR construction cost index for New York
City had a value of 12,381.40 in February 2007.
For Pittsburgh and Atlanta, the values were
7341.32 and 4874.06, respectively. If a general
contractor in Atlanta won construction jobs totaling
$54.3 million, determine their equivalent total
value in New York City.
15.25 The ENR construction cost index (CCI) for August
2010 had a value of 8837.37 when the base year
was 1913 with a value of 100. If the base year is
1967 with a value of 1.0, the CCI for August 2010
would be 8.2272. In this case, what is the CCI
value for 1967 when the base year is 1913?
15.26 The ENR materials cost index (MCI) had a value
of 2708.51 in August 2010. In the same month, the
cost for cement was $96.55 per ton. If cement increased
in price exactly in accordance with the
MCI, what was the cost of cement per ton in 1913
when the MCI index value was 100?
15.27 A contractor purchased equipment costing $40,000
in 2010 when the M&S equipment cost index was
at 1461.3. He remembers purchasing the same
equipment for $21,771 many years ago, but he does
not remember the year that he did so. If the M&S
equipment cost index increased by 2.68% per year
over that time period and the equipment increased
in price exactly in proportion to the index, ( a ) in
what year did he purchase the equipment and ( b )
what was the value of the index in that year?
Cost-Capacity and Factor Methods
15.28 If the materials cost index decreased by 2% over
the same time period that the construction cost
index increased by 2%, what does that suggest
about the labor cost index, if labor represents 35%
of the cost of construction?
15.29 Use the exponent values in Table 15–6 to estimate
the cost for the following equipment to be placed
on an offshore drilling platform.
(a) The cost of a 125-hp centrifugal pump if a
200-hp pump costs $28,000.
(b) The cost of a 1700-gallon stainless steel tank
if a 900-gallon tank costs $4100.
15.30 A high-pressure stainless steel pump (1000 psi)
with a variable-frequency drive (VFD) is installed
in a seawater reverse-osmosis pilot plant that is recovering
water from membrane concentrate at a
rate of 4 gallons per minute (gpm). The cost of the
pump was $13,000. Because of favorable results
from the pilot study, the city utility wants to go with
a full-scale system that will produce 500 gpm.
Determine the estimated cost of the larger pump, if
the exponent in the cost-capacity equation has a
value of 0.37.
15.31 A 0.75 million gallon per day (MGD) induceddraft
packed tower for air-stripping trihalomethanes
from drinking water costs $58,890. Estimate
the cost of a 2-MGD tower if the exponent in the
cost-capacity equation is 0.58.

Problems 407
15.32 The on-site manager told you that the variablefrequency
drive (VFD) for a 300-hp motor costs
$20,000. Make the best cost estimate possible for
the VFD for a 100-hp motor. The exponent in the
cost-capacity equation is not available currently
where you are located.
15.33 The cost of a 68 m 2 falling-film evaporator was
1.52 times the cost of the 30 m 2 unit. What exponent
value in the cost-capacity equation yielded
these results?
15.34 Reinforced concrete pipe (RCP) that is 12 inches
in diameter had a cost of $12.54 per linear foot in
Dallas, Texas in 2007. The cost for 24-inch RCP
was $27.23 per foot. If the cross-sectional area of
the pipe is considered the “ capacity ” in the costcapacity
equation, determine the value of the exponent
in the cost-capacity equation that exactly
relates the two pipe sizes.
15.35 A 100,000 barrel per day (bpd) fractionation tower
cost $1.2 million in 2001 when the Chemical Engineering
plant cost index value was 394.3. How
much would a 450,000 bpd plant cost when the
index value is 575.8, provided the exponent in the
cost-capacity equation is 0.67?
15.36 A mini wind tunnel for calibrating vane or hotwire
anemometers cost $3750 in 2002 when the
M&S equipment index value was 1104.2. If the
index value is now 1620.6, estimate the cost of a
tunnel twice as large. The cost-capacity equation
exponent is 0.89.
15.37 In 2008, a military engineer estimated the cost for
a classified laser-guided device to be $376,900.
The engineer used the M&S equipment index for
the years 1998 and 2008 and the cost-capacity
equation with an exponent value of 0.61. If the
original equipment had only one-fourth the capacity
of the new equipment, what was the cost of the
original equipment in 1998?
15.38 Instead of using a cost-capacity equation for relating
project size and construction cost, the Department
of Defense uses Size Adjustment Factors
(SAFs) that are based on Size Relationship Ratios
(SRRs). For example, a SRR of 2.00 has a SAF of
0.942. Determine the required exponent value in
the cost-capacity equation in order for C 2 to be
0.942 C 1 when Q 2 Q 1 2.00.
15.39 The equipment cost for removing arsenic from a
well that delivers 800 gallons per minute (gpm) is
$1.8 million. If the overall cost factor for this type
of treatment system is 2.25, what is the total plant
cost expected to be?
15.40 A closed-loop filtration system for waterjet cutting
industries eliminates the cost of makeup water
treatments (water softeners, reverse osmosis, etc.)
while maximizing orifice life and machine performance.
If the equipment cost is $225,000 and the
total plant cost is $1.32 million, what is the overall
cost factor for the system?
15.41 The equipment cost for a laboratory that plans to
specialize in analyzing for endocrine disrupters,
pharmaceuticals, and personal care products is
$870,000. If the direct cost factor is 1.32 and the
indirect cost factor is 0.45 (applies to equipment
only), determine the expected cost of the laboratory.
15.42 The equipment cost for a 10 gallon per minute
farm-scale ethanol fuel production plant is
$243,000. The direct cost factor for construction is
1.28 and for installation is 0.23. The indirect cost
factor for licenses, insurance, etc. is 0.84 (applied
to total direct cost). Determine the estimated total
plant cost.
15.43 A chemical engineer at Western Refining has estimated
that the total cost for a diesel fuel desulfurization
system will be $2.3 million. If the direct
cost factor is 1.35 and the indirect cost factor is
0.41, what is the total equipment cost? Both factors
apply to delivered-equipment cost.
15.44 A mechanical engineer estimated that the equipment
cost for a multitube cyclone system with a
capacity of 60,000 cfm would be $400,000. If the
direct cost factor is 3.1 and the indirect cost factor
is 0.38, what is the estimated total plant cost? The
indirect cost factor applies to the total direct cost.
15.45 Nicole is an engineer on temporary assignment at
a refinery operation in Seaside. She has reviewed a
cost estimate for $430,000, which covers some
new processing equipment for the ethylene line.
The equipment itself is estimated at $250,000 with
a construction cost factor of 0.30 and an installation
cost factor of 0.30. No indirect cost factor is
listed, but she knows from other sites that indirect
cost is a sizable amount that increases the cost of
the line’s equipment. ( a ) If the indirect cost factor
should be 0.40, determine whether the current estimate
includes a factor comparable to this value
( b ) Determine the cost estimate if the 0.40 indirect
cost factor is used.
Indirect Cost Allocation
15.46 The company you work for currently allocates insurance
costs on the basis of cost per direct labor
hour. This indirect cost component for the year is
budgeted at $36,000. If the direct labor hours for

408 Chapter 15 Cost Estimation and Indirect Cost Allocation
departments A, B, and C are expected to be 2000,
8000, and 5000, respectively, this year, determine
the allocation to each department.
15.47 The director of public works needs to distribute the
indirect cost allocation of $1.2 million to the three
branches around the city. The recorded amounts
for this year are as follows:
Branch
Miles Driven
Records for This Year
Direct Labor
Hours
North 275,000 38,000
South 247,000 31,000
Midtown 395,000 55,500
The director plans to use the allocation and information
from last year to determine the rates for
this year. This information follows:
Branch
Miles
Driven
Direct
Labor
Hours
Basis
Indirect Cost
Allocation
Last Year, $
North 350,000 40,000 Miles 300,000
South 200,000 20,000 Labor 200,000
Midtown 500,000 64,000 Labor 450,000
(a) Determine the rates for this year for each
branch.
(b) Use the rate to distribute this year’s total indirect
cost. What percentage of this year’s
budget is now distributed?
15.48 A company has a processing department with 10
stations. Because of the nature and use of three of
these stations, each is considered a separate cost
center for indirect cost allocation. The remaining
seven are grouped as one center, CC190. Machine
operating hours are used as the allocation basis for
all machines. A total of $250,000 is allocated to the
department for next year. Use the data collected
this year to determine the indirect cost rate for
each center.
Cost Center
Indirect
Cost Allocated, $
Estimated
Machine Hours
CC100 25,000 800
CC110 50,000 200
CC120 75,000 1200
CC190 100,000 1600
15.49 All the indirect costs are allocated by accounting
for a department. The manager has obtained records
of allocation rates and actual charges for
the prior 3 months and estimates for this month
(May) and next month (see the table). The basis
of allocation is not indicated, and the company
accountant has no record of the basis used. However,
the accountant advises the manager to not
be concerned because the allocation rates have
decreased each month.
Indirect Cost, $
Month Rate Allocated Charged
February 1.40 2800 2600
March 1.33 3400 3800
April 1.37 3500 3500
May 1.03 3600
June 0.92 6000
During the evaluation, the following additional
information from departmental and accounting
records is obtained.
Direct Labor
Month Hours Cost, $
Material
Cost, $
Departmental
Space, ft 2
February 640 2560 5400 2000
March 640 2560 4600 2000
April 640 2560 5700 3500
May 640 2720 6300 3500
June 800 3320 6500 3500
(a) With this information determine the allocation
basis used each month.
(b) Comment on the accountant’s statement
about decreasing allocation rates.
15.50 The mechanical components division manager
asks you to recommend a make/buy decision on a
major automotive subassembly that is currently
purchased externally for a total of $3.9 million
this year. This cost is expected to continue rising
at a rate of $300,000 per year. Your manager asks
that both direct and indirect costs be included
when in-house manufacturing (make alternative)
is evaluated. New equipment will cost $3 million
and will have a salvage of $0.5 million and a life
of 6 years. Estimates of materials, labor costs, and
other direct costs are $1.5 million per year. Typical
indirect rates, bases, and expected usage are
shown below. Perform the AW evaluation at
MARR 12% per year over a 6-year study period.
Show both hand and spreadsheet solutions,
as directed.
Department Basis Rate
Expected
Usage
X Direct labor cost $2.40 per $ $450,000
Y Materials cost $0.50 per $ $850,000
Z Number of
inspections
$20 per
inspection
4,500

Problems 409
ABC Method
15.51 The municipal water and desalinization utility in a
California city currently allocates some costs for
maintenance shop workers to pumping stations
based on the number of pumps at each station. At
the last director’s semiannual meeting, a suggestion
was made to change the allocation basis to the
number of trips that pump service personnel make
to each station, because some stations have old
pumps that require more maintenance. Information
about the stations is below. The indirect cost
budget is $20,000 per pump.
( a) Allocate the budget to each station based on
the number of service trips.
( b ) Determine the old allocation on the basis of
the number of pumps, and comment on any
significant differences in the amounts allocated
to the stations.
Station ID No. of Pumps Service Trips per Year
Sylvester 5 190
Laurel 7 55
7 th St 3 38
Spicewood 4 104
15.52 Factory Direct manufactures and sells manufactured
homes. Traditionally, it has distributed indirect costs
to its three construction plants based on materials
cost. Each plant builds different models and floor
plans. Advances in weight and shape of plastic and
wood composite components have decreased cost
and time to produce a unit. Because of these advances,
the CFO plans to use build-time per unit as
the new basis. However, he initially wants to determine
what the allocation would have been this year
had build-time been the basis prior to incorporation
of the new materials. The data shown represents average
costs and times. Use this data and the three
bases indicated to determine the allocation rates and
indirect cost distribution of $900,000 for this year.
Plant Texas Oklahoma Kansas
Direct material cost,
$ per unit
Previous build-time
per unit, work-hours
New build-time per
unit, work-hours
20,000 12,700 18,600
400 415 580
425 355 480
Problems 15.53 through 15.55 use the following
information.
Jet Green Airways historically distributes the indirect
costs of lost and damaged baggage to its three major hubs
using a basis of annual number of flights in and out of
each hub. Last year $667,500 was distributed as follows:
Hub Airport Flights Rate, $ per Flight Allocation, $
DFW 55,000 6 330,000
YYZ 20,833 9 187,500
MEX 15,000 10 150,000
The airline’s baggage management director suggests that
an allocation on the basis of baggage traffic, not flights,
will better represent the distribution, primarily based on
the fact that the high fees now charged to passengers to
check luggage have significantly changed the number of
bags handled at the major hubs. Total number of bags
handled during the year are 2,490,000 at DFW, 1,582,400
at YYZ, and 763,500 at MEX.
15.53 What are the activity and the cost driver for the
suggested baggage-traffic basis?
15.54 Using the baggage-traffic basis, determine the allocation
rate using last year’s total of $667,500,
and distribute this amount to the hubs this year.
15.55 What are the percentage changes in allocation at
each hub using the two different bases?
15.56 On-line Vacation distributes advertising costs to its
four resort sites in the Caribbean on the basis of
the size of the resort budget. For this year, in round
numbers, the budgets and allocation of $1 million
advertising indirect costs are as follows:
Site
A B C D
Budget, $ 2 million 3 million 4 million 1 million
Allocation, $ 200,000 300,000 400,000 100,000
(a)
(b)
Determine the allocation if the ABC method is
used with a new basis. Define the activity as the
advertising department at each resort. The cost
driver is the number of guests during the year.
Site
A B C D
Guests 3500 4000 8000 1000
Again use the ABC method, but now make the
cost driver the total number of guest nights at
each resort. The average number of lodging
nights for guests at each site is as follows:
Site
A B C D
Length of stay, nights 3.0 2.5 1.25 4.75
(c) Comment on the distribution of advertising
costs using the two methods. Identify any
other cost drivers that might be considered
for the ABC approach that may reflect a
realis tic allocation of the costs.

410 Chapter 15 Cost Estimation and Indirect Cost Allocation
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
15.57 In the bottom-up approach to cost estimating:
(a) Required price is an input variable.
(b) Cost estimates are an output variable.
(c) Required price is an output variable.
(d) Both ( a ) and ( b ) are correct.
15.58 A ratio of the cost of something today to its cost at
some time in the past is called a:
(a) Cost-capacity index
(b) Cost index
(c) Buyer’s guide
(d) Bluebook index
15.59 Index values could probably be obtained from all
of the following places except:
(a) Trade organizations
(b) Engineering News Record magazine
(c) Government organizations
(d) Home Depot or Lowe’s home improvement
stores
15.60 A 50-hp turbine pump was purchased for $2100. If
the exponent in the cost-capacity equation has a
value of 0.76, a 200-hp turbine pump could be expected
to cost about:
( a ) $6020
( b ) $5320
( c ) $4890
( d ) $4260
100,000 units per day was $3 million, the value of
the exponent in the cost-capacity equation is:
( a ) 0.26
( b ) 0.39
( c ) 0.45
( d ) 0.60
15.64 The equipment for applying specialty coatings
that provide a high angle of skid for the paperboard
and corrugated box industries has a delivered
cost of $390,000. If the overall cost factor
for the complete system is 2.96, the total plant
cost is approximately:
( a ) $954,400
( b ) $1,054,400
( c ) $1,154,400
( d ) $1,544,400
15.65 The delivered-equipment cost for setting up a
production and assembly line for high-sensitivity,
gas-damped accelerometers is $650,000. If the direct
cost and indirect cost factors are 1.82 and
0.31, respectively, and both factors apply to
delivered-equipment cost, the total plant cost estimate
is approximately:
( a ) $2,034,500
( b ) $1,734,500
( c ) $1,384,500
( d ) $1,183,000
15.61 The city built a recreation park in 1980 for
$500,000. The ENR construction cost index had
a value of 3378.17 at that time. If the city is
planning to construct a similar recreation park
when the index value is 5542.16, the estimated
cost is closest to:
( a ) $695,800
( b ) $750,700
( c ) $820,300
( d ) $910,500
15.62 A small company bought a 250-hp compressor in
1998 for $3000 when the M&S equipment cost
index had a value of 1061.9. If the exponent in the
cost-capacity equation is 0.32 and the M&S index
value was 1449.3 in 2008, the cost of a 500-hp
compressor in 2008 was closest to:
( a ) $3744
( b ) $4094
( c ) $4627
( d ) $5110
15.63 The cost for implementing a manufacturing process
that has a capacity of 6000 units per day was
$550,000. If the cost for a plant with a capacity of
15.66 A police department wants to allocate the indirect
cost of speed monitoring to the three toll roads
around the city. An allocation basis that may not be
reasonable is:
(a) Miles of toll road monitored
(b) Average number of cars patrolling per hour
( c) Amount of car traffic per section of toll
road
(d) Cost to operate a patrol car
15.67 The IT department allocates indirect costs to user
departments on the basis of CPU time at the rate of
$2000 per second. For the first quarter, the two
heaviest-use departments logged 900 and 1300 seconds,
respectively. If the IT indirect budget for the
year is $8.0 million, the percentage of this year’s
allocation consumed by these departments is
closest to:
( a ) 32%
( b ) 22.5%
( c ) 55%
( d ) Not enough information to determine
15.68 If the engineering department is the activity to
receive indirect cost allocation for the year, cost

Case Study 411
drivers for the ABC method that seem reasonable
may be:
1. Cost of engineering changes processed
2. Size of the workforce
3. Administrative cost to process a change
order
( a ) 1
( b ) 2
( c ) 3
( d ) 1 and 3
15.69 Advantages of the ABC method are:
1. It is an excellent replacement for a traditional
cost accounting system.
2. It is always cheaper to operate than a traditional
allocation system.
3. It can help explain the economic impact of
management decisions.
( a ) 1
( b ) 2
( c ) 3
( d ) 1 and 3
CASE STUDY
INDIRECT COST ANALYSIS OF MEDICAL EQUIPMENT MANUFACTURING COSTS
Background
all its other product lines. The same was applied when Quik-
Sterz was priced. However, Arnie, the person who performed
Three years ago Medical Dynamics, a medical equipment
the indirect cost analysis and set the sales price, is no longer
unit of Johnson and Sons, Inc., initiated the manufacture and
at the company, and the detailed analysis is no longer available.
Through e-mail and telephone conversations, Arnie said
sales of a portable sterilization unit (Quik-Sterz) that can be
placed in the hospital room of a patient. This unit sterilizes
the current price was set at about 10% above the total manufacturing
cost determined 2 years ago, and that some records
and makes available at the bedside some of the reusable instruments
that nurses and doctors usually obtain by walking
were available in the design department files. A search of
to or receiving delivery from a centralized area. This new unit
these files revealed the manufacturing and cost information in
makes the instruments available at the point and time of use
Table 15–11. It is clear from these and other records that
for burn and severe wound patients who are in a regular patient
room.
Arnie used traditional indirect cost analysis based on direct
labor hours to estimate the total manufacturing costs of
There are two models of Quik-Sterz sold. The standard
$9.73 per unit for the standard model and $27.07 per unit for
version sells for $10.75, and a premium version with customized
trays and a battery backup system sells for $29.75. The
the premium model.
Last year management decided to place the entire plant on
product has sold well to hospitals, convalescent units, and
the ABC system of indirect cost allocation. The costs and
nursing homes at the level of about 1 million units per year.
sales figures collected for Quik-Sterz the year before were
still accurate. Five activities and their cost drivers were identified
for the Medical Dynamics manufacturing operations
Information
Medical Dynamics has historically used an indirect cost allocation
system based upon direct hours to manufacture for marized in this
(Table 15–12). Also, the volumes for each model are sum-
table.
TABLE 15–11
Historical Records of Direct and Indirect Cost Analyses for Quik-Sterz
Quik-Sterz Direct Cost (DC) Evaluation
Direct Material,
$Unit
Model
Direct Labor,
$Unit 1
Direct Labor,
HoursUnit
Total Direct
Labor Hours
Standard 5.00 2.50 0.25 187,500
Premium 10.00 3.75 0.50 125,000
Quik-Sterz Indirect Cost (IDC) Evaluation
Fraction IDC
Allocated
Model
Direct Labor,
HoursUnit
Allocated
IDC, $
Sales,
Units/Year
Standard 0.25 —
3
1
Premium 0.50 — 2 3
1.67 million 750,000
3.33 million 250,000
1 Average direct labor rate is $20 per hour.

412 Chapter 15 Cost Estimation and Indirect Cost Allocation
TABLE 15–12
Quik-Sterz Activities, Cost Drivers, and Volume Levels for ABC-Based
Indirect Cost Allocation
Activity Cost Driver VolumeYear
Actual
Cost, $/Year
Quality Inspections 20,000 inspections 800,000
Purchasing Purchase orders 40,000 orders 1,200,000
Scheduling Change orders 1,000 orders 800,000
Production setup Setups 5,000 setups 1,000,000
Machine operations Machine hours 10,000 hours 1,200,000
Volume Level for the Year
Cost Driver Standard Premium
Quality inspections 8,000 12,000
Purchase orders 30,000 10,000
Number of change orders 400 600
Production setup 1,500 3,500
Machine hours 7,000 3,000
The ABC method will be used henceforth, with the intention
of determining the total cost and price based on its results. The
first impression of the production manager is that the new system
will show that indirect costs for Quik-Sterz are about the
same as they have been for other products over the last several
years when a standard model and an upgrade (premium) model
were sold. Predictably, they state, the standard model will receive
about 13 of the indirect cost, and the premium will receive
the remaining 23. Fundamentally, there are two reasons
why the production manager does not like to produce premium
versions: They are less profitable for the company, and they require
significantly more time and operations to manufacture.
Case Study Exercises
1. Use traditional indirect cost allocation to verify Arnie’s
cost and price estimates.
2. Use the ABC method to estimate the indirect cost allocation
and total cost for each model.
3. If the prices and number of units sold are the same
next year (750,000 standard and 250,000 premium),
and all other costs remain constant, compare the
profit from Quik-Sterz under the ABC method with
the profit using the traditional indirect cost allocation
method.
4. What prices should Medical Dynamics charge next year
based on the ABC method and a 10% markup over cost?
What is the total profit from Quik-Sterz predicted to be
if sales hold steady?
5. Using the results above, comment on the production
manager’s prediction of indirect costs using ABC (13
standard; 23 premium) and the two reasons given to
not produce the premium version of Quik-Sterz.
CASE STUDY
DECEPTIVE ACTS CAN GET YOU IN TROUBLE
Contributed by Dr. Paul Askenasy, Agronomist, Texas Commission
on Environmental Quality
Background
Surface mining of coal is the removal of soil and sediments
from underlying strata that lie above the material to be mined.
The law requires that land disturbed by these types of mining
activities be returned to a productive capacity that is as good
as or better than its productive capacity before mining.
The productive capacity of soils is directly correlated to
the textural (sand, silt, and clay content) and chemical
characteristics of the soil (for example, pH). To this end,
mining companies must sample the different soils found in
the areas to be disturbed by mining activities. The purpose
of the sampling is to establish a baseline characterizing the
textural and chemical makeup of the soils prior to mining.
Soils in a low pH range (pH values 5) are indicative of
low fertility. Once the natural resource, such as coal, is removed,
the pit is backfilled with sediments and the terrain
surface is contoured to reestablish the premine drainages.
To meet the baseline for pH, the acreage of the mine soils
with low pH should not exceed the acreage of the unmined
soils with low pH.

Case Study 413
Information
Yucatan Mining Company (not the actual name) planned to
disturb 600 acres due to mining activities. The different soils
within the 600 acres were depicted in the County Soil Survey
where the mining activities were to take place. Prior to mining,
the company obtained soil samples from 10 different locations
within each soil type and had them analyzed for a
number of parameters including pH. Assessment of the data
indicated that 30% of the area (180 acres) occupied by the
soils in the area to be disturbed had pH values between 4.0
and 4.9. The application for mining was approved by the
State Department of Mining and Reclamation.
Six years later, 450 acres had been mined and the terrain
surface had been leveled to reestablish premine slopes. Of the
450 acres leveled, 175 acres had pH values between 4.0 and
4.9. The president of Yucatan indicated that the company
would submit a revised soil baseline based on new sampling
in the remaining 150 acres of unmined soils because, in his
opinion, the first soil baseline was biased.
The request to do more soil analyses to augment their existing
soil baseline was approved. The company quickly hired
a consultant to develop the new baseline, and subsequently
Yucatan submitted the final report from its consultant to the
State Department of Mining and Reclamation. This revised
premine soil baseline indicated that 45% of the premine soils
had pH values between 4.0 and 4.9. Comparative results between
the old and new samples can be expressed as follows:
Percent and Acreage of Area
Soil Baselines Old Soil Baseline Revised Soil Baseline
pH: 4.0 4.9 30% 45%
180 acres 270 acres
A rough statistical check between the old and revised soil
baselines indicated that the results were mixed. Based on this
preliminary result and the fact that there was a significant increase
in the percent of area with low-productivity soil, an
in-depth analysis of the revised baseline sample study was
performed. Contained in the submitted new-sample package
was a letter from the Yucatan consultant. It indicated to
Yucatan’s management that 100 separate soil samples had
been obtained and analyzed and that the revised premine soil
baseline had been developed using the data from the 30 samples
with the lowest pH values .
The State Department of Mining and Reclamation staff
concluded that the revised soil pH sample data had been carefully
“screened” to reduce the amount of remediation work
that Yucatan Mining would have to complete . Within a week,
Yucatan was notified that further review of the revised soil
baseline could not be pursued, because it appeared the revised
soil baseline was developed using a technique that
skewed results in favor of lower pH values. It was also noted
that should Yucatan Mining disagree with this response, the
case would be filed with the legal staff as a contested case.
Within several days, the Yucatan president responded indicating
that the company was withdrawing the new application
from consideration by the department.
Case Study Questions
1. Assume you are the director of the State Department of
Mining and Reclamation and were informed of the findings
on the new samples versus the old samples. What
actions would you direct your staff to take concerning
this situation?
2. Suppose there had been several cases of deceptive acts
similar to this one over the last few years. What type of
“audit” procedures might you want implemented to
identify these possibly unethical activities?
3. Yucatan clearly would state that both the old and new
samples were randomly located about the entire mining
area. When the 30 lowest pH samples were used to establish
the new baseline, were the samples still random,
according to experimental design standards? If so, why?
If not, why not?
4. You and the president of Yucatan Mining have been acquaintances
for some years. You have golfed together
several times, your and his children are on the same soccer
team at school, and your families are members of
the same community swimming pool club. What effect
would this event have upon your and your family’s relationships
with the family of the Yucatan president? How
would you handle this situation?
5. As a matter of principle and practice, do you believe
there is some amount of data-altering or bias-making
that is allowed before an application (such as the one
described here) should be considered the result of professionally
unethical acts? How would you define such
a threshold limit?

CHAPTER 16
Depreciation
Methods
LEARNING OUTCOMES
Purpose: Use depreciation or depletion methods to reduce the book value of a capital investment in an asset and natural
resource.
SECTION TOPIC LEARNING OUTCOME
16.1 Terminology • Define and use the basic terms of asset
depreciation.
16.2 Straight line • Apply the straight line (SL) method of
depreciation.
16.3 Declining balance • Apply the declining balance (DB) and double
declining balance (DDB) methods of
depreciation.
16.4 MACRS • Apply the modified accelerated cost recovery
system for tax depreciation purposes for U.S.-
based corporations.
16.5 Recovery period • Select the asset recovery period for MACRS
depreciation.
16.6 Depletion • Explain depletion; apply cost depletion and
percentage depletion methods.
Chapter 16 Appendix
16A.1 Historical methods • Apply the sum-of-years-digits (SYD) and unitof-production
(UOP) methods of depreciation.
16A.2 Switching • Switch between classical depreciation methods;
explain how MACRS provides for switching.
16A.3 MACRS and switching • Calculate MACRS rates using switching between
classical methods and MACRS rules.

T
he capital investments of a corporation in tangible assets—equipment, computers,
vehicles, buildings, and machinery—are commonly recovered on the books
of the corporation through depreciation . Although the depreciation amount is
not an actual cash flow, the process of depreciating an asset on the books of the corporation
accounts for the decrease in an asset’s value because of age, wear, and obsolescence.
Even though an asset may be in excellent working condition, the fact that it is worth less
through time is taken into account in after-tax economic evaluation studies. An introduction
to depreciation types, terminology, and classical methods is followed by a discussion
of the Modified Accelerated Cost Recovery System (MACRS) , which is the standard in the
United States for tax purposes. Other countries commonly use the classical methods for tax
computations.
Why is depreciation important to engineering economy? Depreciation is a tax-allowed
deduction included in tax calculations in virtually all industrialized countries. Depreciation
lowers income taxes via the general relation
Taxes (income deductions)(tax rate)
Income taxes are discussed further in Chapter 17.
This chapter concludes with an introduction to two methods of depletion, which are used
to recover capital investments in deposits of natural resources such as oil, gas, minerals, ores,
and timber.
The chapter appendix describes two historically useful methods of depreciation—
sum-of-years-digits and unit-of-production. Additionally, the appendix includes an in-depth
derivation of the MACRS depreciation rates from the straight line and declining balance
rates. This is accomplished using a procedure called switching between classical depreciation
methods.
16.1 Depreciation Terminology
The concept and types of depreciation are defined here. Most descriptions are applicable to corporations
as well as individuals who own depreciable assets.
Depreciation is a book method (noncash) to represent the reduction in value of a tangible asset.
The method used to depreciate an asset is a way to account for the decreasing value of the asset
to the owner and to represent the diminishing value (amount) of the capital funds invested in it.
The annual depreciation amount is not an actual cash flow , nor does it necessarily reflect the
actual usage pattern of the asset during ownership.
Though the term amortization is sometimes used interchangeably with the term depreciation,
they are different. Depreciation is applied to tangible assets, while amortization is used to reflect
the decreasing value of intangibles, such as loans, mortgages, patents, trademarks, and goodwill.
In addition, the term capital recovery is sometimes used to identify depreciation. This is clearly
a different use of the term than what we learned in Chapter 5. The term depreciation is used
throughout this book.
There are two different purposes for using the depreciation methods we will cover in this
chapter:
Book depreciation Used by a corporation or business for internal financial accounting to
track the value of an asset or property over its life.
Tax depreciation Used by a corporation or business to determine taxes due based on current
tax laws of the government entity (country, state, province, etc.). Even though depreciation
itself is not a cash flow, it can result in actual cash flow changes because the amount of
tax depreciation is a deductible item when calculating annual income taxes for the corporation
or business.
The methods applied for these two purposes may or may not utilize the same formulas. Book
depreciation indicates the reduced investment in an asset based upon the usage pattern and
expected useful life of the asset. There are classical, internationally accepted depreciation methods
used to determine book depreciation: straight line, declining balance, and the historical

416 Chapter 16 Depreciation Methods
sum-of-years-digits method. The amount of tax depreciation is important in an after-tax engineering
economy study and will vary among nations.
In most industrialized countries, the annual tax depreciation is tax-deductible ; that is, it is subtracted
from income when calculating the amount of taxes due each year. However, the tax
depreciation amount must be calculated using a government-approved method.
Tax depreciation may be calculated and referred to differently in countries outside the United
States. For example, in Canada the equivalent is CCA (capital cost allowance), which is calculated
based on the undepreciated value of all corporate properties that form a particular
class of assets, whereas in the United States depreciation may be determined for each asset
separately.
Where allowed, tax depreciation is usually based on an accelerated method , whereby the
depreciation for the first years of use is larger than that for later years. In the United States this
method is called MACRS, as covered in later sections. In effect, accelerated methods defer
some of the income tax burden to later in the asset’s life; they do not reduce the total tax
burden.
Common terms used in depreciation are explained here.
First cost P or unadjusted basis B is the delivered and installed cost of the asset including
purchase price, delivery and installation fees, and other depreciable direct costs incurred to
prepare the asset for use. The term unadjusted basis , or simply basis , is used when the asset is
new, with the term adjusted basis used after some depreciation has been charged. When the
first cost has no added, depreciable costs, the basis is the first cost, that is, P B .
Book value BV t represents the remaining, undepreciated capital investment on the books after
the total amount of depreciation charges to date has been subtracted from the basis. The book
value is determined at the end of each year t ( t 1, 2, . . . , n ), which is consistent with the
end-of-year convention.
Recovery period n is the depreciable life of the asset in years. Often there are different
n values for book and tax depreciation. Both of these values may be different from the asset’s
estimated productive life.
Market value MV , a term also used in replacement analysis, is the estimated amount realizable
if the asset were sold on the open market. Because of the structure of depreciation laws,
the book value and market value may be substantially different. For example, a commercial
building tends to increase in market value, but the book value will decrease as depreciation
charges are taken. However, a computer workstation may have a market value much lower
than its book value due to rapidly changing technology.
Salvage value S is the estimated trade-in or market value at the end of the asset’s useful life.
The salvage value, expressed as an estimated dollar amount or as a percentage of the first cost,
may be positive, zero, or negative due to dismantling and carry-away costs.
Depreciation rate or recovery rate d t is the fraction of the first cost removed by depreciation
each year t. This rate may be the same each year, which is called the straight line rate d , or
different for each year of the recovery period.
Personal property, one of the two types of property for which depreciation is allowed is
the income-producing, tangible possessions of a corporation used to conduct business.
Included is most manufacturing and service industry property—vehicles, manufacturing
equipment, materials handling devices, computers and networking equipment, communications
equipment, office furniture, refining process equipment, construction assets, and
much more.
Real property includes real estate and all improvements—office buildings, manufacturing
structures, test facilities, warehouses, apartments, and other structures. Land itself is considered
real property, but it is not depreciable.
Half-year convention assumes that assets are placed in service or disposed of in midyear,
regardless of when these events actually occur during the year. This convention is utilized in
this text and in most U.S.-approved tax depreciation methods. There are also midquarter and
midmonth conventions.

16.1 Depreciation Terminology 417
Figure 16–1
General shape of book
value curves for different
depreciation methods.
SL
Book value, $
MACRS
DB
Estimated
salvage
value
Time, years
Recovery period
As mentioned before, there are several models for depreciating assets. The straight line (SL)
method is used historically and internationally. Accelerated models, such as the declining bal -
ance (DB) method, decrease the book value to zero (or to the salvage value) more rapidly than
the straight line method, as shown by the general book value curves in Figure 16–1.
For each of the methods—straight line, declining balance, MACRS, and sum-of-yearsdigits—there
are spreadsheet functions available to determine annual depreciation. Each function
is introduced and illustrated as the method is explained.
As expected, there are many rules and exceptions to the depreciation laws of a country. One
that may be of interest to a U.S.-based small or medium-sized business performing an economic
analysis is the Section 179 Deduction. This is an economic incentive that changes over the years
and encourages businesses to invest capital in equipment directly used in the company. Up to a
specified amount, the entire basis of an asset is treated as a business expense in the year of purchase.
This tax treatment reduces federal income taxes, just as depreciation does, but it is allowed
in lieu of depreciating the first cost over several years. The limit changes with time; it was
$24,000 in 2002; $102,000 in 2004; $125,000 in 2007; and $250,000 in 2008–2010. The economic
stimulus efforts in the United States and around the world during the latter part of the decade
made many attempts to put investment capital to work in small and medium-sized businesses.
Investments above these limits must be depreciated using MACRS.
In the 1980s the U.S. government standardized accelerated methods for federal tax depreciation
purposes. In 1981, all classical methods, including straight line, declining balance, and sum-of-yearsdigits
depreciation, were disallowed as tax deductible and replaced by the Accelerated Cost Recovery
System (ACRS). In a second round of standardization, MACRS (Modified ACRS) was made the
required tax depreciation method in 1986. To this date, the following is the law in the United States.
Tax depreciation must be calculated using MACRS; book depreciation may be calculated
using any classical method or MACRS.
MACRS has the DB and SL methods, in slightly different forms, embedded in it, but these two
methods cannot be used directly if the annual depreciation is to be tax deductible. Many U.S.
companies still apply the classical methods for keeping their own books, because these methods
are more representative of how the usage patterns of the asset reflect the remaining capital invested
in it. Most other countries still recognize the classical methods of straight line and declining
balance for tax or book purposes. Because of the continuing importance of the SL and DB
methods, they are explained in the next two sections prior to MACRS. Appendix Section 16A.1
discusses two historical methods of depreciation.

418 Chapter 16 Depreciation Methods
Tax law revisions occur often, and depreciation rules are changed from time to time in the
United States and other countries. For more depreciation and tax law information, consult the
U.S. Department of the Treasury, Internal Revenue Service (IRS), website at www.irs.gov. Pertinent
publications can be downloaded. Publication 946, How to Depreciate Property, is especially
applicable to this chapter. MACRS and most corporate tax depreciation laws are discussed in it.
16.2 Straight Line (SL) Depreciation
Straight line depreciation derives its name from the fact that the book value decreases linearly
with time. The depreciation rate is the same (1 n ) each year of the recovery period n .
Straight line depreciation is considered the standard against which any depreciation model is
compared. For book depreciation purposes, it offers an excellent representation of book value for
any asset that is used regularly over an estimated number of years. For tax depreciation, as mentioned
earlier, it is not used directly in the United States, but it is commonly used in most countries
for tax purposes. However, the U.S. MACRS method includes a version of SL depreciation
with a larger n value than that prescribed by regular MACRS (see Section 16.5).
The annual SL depreciation is determined by multiplying the first cost minus the salvage value
by d t . In equation form,
D t ( B S ) d t
B S ———
n [16.1]
Book Value
B
S
Time
1
where t year ( t 1, 2, . . . , n )
D t annual depreciation charge
B first cost or unadjusted basis
S estimated salvage value
n recovery period
d t depreciation rate 1 n
Since the asset is depreciated by the same amount each year, the book value after t years of service,
denoted by BV t , will be equal to the first cost B minus the annual depreciation times t .
BV t B tD t [16.2]
Earlier we defined d t as a depreciation rate for a specific year t . However, the SL model has the
same rate for all years, that is,
d d t — 1 n [16.3]
The format for the spreadsheet function to display the annual depreciation D t in a single-cell
operation is
SLN( B , S , n) [16.4]
EXAMPLE 16.1
If an asset has a first cost of $50,000 with a $10,000 estimated salvage value after 5 years,
( a ) calculate the annual depreciation and ( b ) calculate and plot the book value of the asset after
each year, using straight line depreciation.
Solution
(a) The depreciation each year for 5 years can be found by Equation [16.1].
D t B S ———
n
50,000 10,000
——————— $8000
5
Enter the function SLN(50000,10000,5) in any cell to display the D t of $8000.

16.3 Declining Balance (DB) and Double Declining Balance (DDB) Depreciation 419
Book value BV t $1000
50
40
30
20
10
D t = $8000
Figure 16–2
Book value of an asset
using straight line depreciation,
Example 16.1.
S = $10,000
0 1 2 3
Year t
4 5
t
(b) The book value after each year t is computed using Equation [16.2]. The BV t values are
plotted in Figure 16–2. For years 1 and 5, for example,
BV 1 50,000 1(8000) $42,000
BV 5 50,000 5(8000) $10,000 S
16.3 Declining Balance (DB) and Double Declining
Balance (DDB) Depreciation
The declining balance method is commonly applied as the book depreciation method. Like the
SL method, DB is embedded in the MACRS method, but the DB method itself cannot be used to
determine the annual tax-deductible depreciation in the United States. This method is used routinely
in most other countries for tax and book depreciation purposes.
Declining balance is also known as the fixed percentage or uniform percentage method.
DB depreciation accelerates the write-off of asset value because the annual depreciation is determined
by multiplying the book value at the beginning of a year by a fixed (uniform) percentage d ,
expressed in decimal form. If d 0.1, then 10% of the book value is removed each year. Therefore,
the depreciation amount decreases each year.
The maximum annual depreciation rate for the DB method is twice the straight line rate, that is,
d max 2 n [16.5]
Book Value
B
Time
In this case the method is called double declining balance (DDB) . If n 10 years, the DDB rate
is 210 0.2; so 20% of the book value is removed annually. Another commonly used percentage
for the DB method is 150% of the SL rate, where d 1.5 n .
The depreciation for year t is the fixed rate d times the book value at the end of the previous year.
D t ( d )BV t 1 [16.6]
The actual depreciation rate for each year t , relative to the basis B , is
d t d (1 d ) t 1 [16.7]
If BV t 1 is not known, the depreciation in year t can be calculated using B and d .
D t dB (1 d ) t1 [16.8]
Book value in year t is determined in one of two ways: by using the rate d and basis B or by
subtracting the current depreciation charge from the previous book value. The equations are
BV t B (1 d ) t [16.9]
BV t BV t 1 D t [16.10]

420 Chapter 16 Depreciation Methods
It is important to understand that the book value for the DB method never goes to zero, because
the book value is always decreased by a fixed percentage. The implied salvage value after n
years is the BV n amount, that is,
Implied S BV n B (1 d ) n [16.11]
If a salvage value is estimated for the asset, this estimated S value is not used in the DB or
DDB method to calculate annual depreciation. However, if the implied S estimated S , it is
necessary to stop charging further depreciation when the book value is at or below the estimated
salvage value. In most cases, the estimated S is in the range of zero to the implied S value. (This
guideline is important when the DB method can be used directly for tax depreciation purposes.)
If the fixed percentage d is not stated, it is possible to determine an implied fixed rate using the
estimated S value, if S 0. The range for d is 0 d 2 n .
Implied d 1 ( S —
B ) 1n [16.12]
The spreadsheet functions DDB and DB are used to display depreciation amounts for specific
years. The function is repeated in consecutive spreadsheet cells because the depreciation amount
D t changes with t . For the double declining balance method, the format is
DDB( B , S , n , t , d ) [16.13]
The entry d is the fixed rate expressed as a number between 1 and 2. If omitted, this optional
entry is assumed to be 2 for DDB. An entry of d 1.5 makes the DDB function display 150%
declining balance method amounts. The DDB function automatically checks to determine when
the book value equals the estimated S value. No further depreciation is charged when this
occurs. (To allow full depreciation charges to be made, ensure that the S entered is between zero
and the implied S from Equation [16.11].) Note that d 1 is the same as the straight line rate
1 n , but D t will not be the SL amount because declining balance depreciation is determined as a
fixed percentage of the previous year’s book value, which is completely different from the SL
calculation in Equation [16.1].
The format for the DB function is DB( B , S , n , t ). Caution is needed when using this function.
The fixed rate d is not entered in the DB function; d is an embedded calculation using
a spreadsheet equivalent of Equation [16.12]. Also, only three significant digits are maintained
for d , so the book value may go below the estimated salvage value due to round-off
errors. Therefore, if the depreciation rate is known, always use the DDB function to ensure
correct results. Examples 16.2 and 16.3 illustrate DB and DDB depreciation and their spreadsheet
functions.
EXAMPLE 16.2
Underwater electroacoustic transducers were purchased for use in SONAR applications. The
equipment will be DDB depreciated over an expected life of 12 years. There is a first cost of
$25,000 and an estimated salvage of $2500. ( a ) Calculate the depreciation and book value for
years 1 and 4. Write the spreadsheet functions to display depreciation for years 1 and 4.
( b ) Calculate the implied salvage value after 12 years.
Solution
(a) The DDB fixed depreciation rate is d 2 n 212 0.1667 per year. Use Equations [16.8]
and [16.9].
Year 1: D 1 (0.1667)(25,000)(1 0.1667) 11 $4167
BV 1 25,000(1 0.1667) 1 $20,833
Year 4: D 4 (0.1667)(25,000)(1 0.1667) 41 $2411
BV 4 25,000(1 0.1667) 4 $12,054
The DDB functions for D 1 and D 4 are, respectively, DDB(25000,2500,12,1) and
DDB(25000,2500,12,4).

16.3 Declining Balance (DB) and Double Declining Balance (DDB) Depreciation 421
(b) From Equation [16.11], the implied salvage value after 12 years is
Implied S 25,000(1 − 0.1667) 12 $2803
Since the estimated S $2500 is less than $2803, the asset is not fully depreciated when
its 12-year expected life is reached.
EXAMPLE 16.3
Freeport-McMoRan Copper and Gold has purchased a new ore grading unit for $80,000. The
unit has an anticipated life of 10 years and a salvage value of $10,000. Use the DB and DDB
methods to compare the schedule of depreciation and book values for each year. Solve by hand
and by spreadsheet.
Solution by Hand
An implied DB depreciation rate is determined by Equation [16.12].
d 1 ( 10,000 ———
80,000 ) 110 0.1877
Note that 0.1877 2 n 0.2, so this DB model does not exceed twice the straight line rate.
Table 16–1 presents the D t values using Equation [16.6] and the BV t values from Equation
[16.10] rounded to the nearest dollar. For example, in year t 2, the DB results are
D 2 d (BV 1 ) 0.1877(64,984) $12,197
BV 2 64,984 12,197 $52,787
Because we round off to even dollars, $2312 is calculated for depreciation in year 10, but
$2318 is deducted to make BV 10 S $10,000 exactly. Similar calculations for DDB with
d 0.2 result in the depreciation and book value series in Table 16–1.
TABLE 16–1 D t and BV t Values for DB and DDB Depreciation, Example 16.3
Declining Balance, $ Double Declining Balance, $
Year t D t BV t D t BV t
0 — 80,000 — 80,000
1 15,016 64,984 16,000 64,000
2 12,197 52,787 12,800 51,200
3 9,908 42,879 10,240 40,960
4 8,048 34,831 8,192 32,768
5 6,538 28,293 6,554 26,214
6 5,311 22,982 5,243 20,972
7 4,314 18,668 4,194 16,777
8 3,504 15,164 3,355 13,422
9 2,846 12,318 2,684 10,737
10 2,318 10,000 737 10,000
Solution by Spreadsheet
The spreadsheet in Figure 16–3 displays the results for the DB and DDB methods. The chart
plots book values for each year. Since the fixed rates are close—0.1877 for DB and 0.2 for
DDB—the annual depreciation and book value series are approximately the same for the two
methods.
The depreciation rate (cell B5) is calculated by Equation [16.12], but note in the cell tags
that the DDB function is used in both columns B and D to determine annual depreciation.
As mentioned earlier, the DB function automatically calculates the implied rate by Equation
[16.12] and maintains it to only three significant digits. Therefore, if the DB function

422 Chapter 16 Depreciation Methods
1(B3/B2)^0.1
DDB(B$2,B$3,B$4,$A18,10*$B$5)
DDB(B$2,B$3,B$4,$A18)
Figure 16–3
Annual depreciation and book value using DB and DDB methods, Example 16.3.
were used in column B (Figure 16–3), the fixed rate applied would be 0.188. The resulting
D t and BV t values for years 8, 9, and 10 would be as follows:
t D t , $ BV t , $
8 3,501 15,120
9 2,842 12,277
10 2,308 9,969
Also noteworthy is the fact that the DB function uses the implied rate without a check to halt
the book value at the estimated salvage value. Thus, BV 10 will go slightly below S $10,000,
as shown above. However, the DDB function uses a relation different from that of the DB function
to determine annual depreciation—one that correctly stops depreciating at the estimated
salvage value, as shown in Figure 16–3, cells E17–E18.
16.4 Modified Accelerated Cost Recovery
System (MACRS)
In the 1980s, the United States introduced MACRS as the required tax depreciation method for
all depreciable assets. Through MACRS, the 1986 Tax Reform Act defined statutory depreciation
rates that take advantage of the accelerated DB and DDB methods. Corporations are free to apply
any of the classical methods for book depreciation. When developed, MACRS and its predecessor
ACRS were intended to create economic growth through the investment of new capital and
the tax advantages that accelerated depreciation methods offer corporations and businesses. 1
Many aspects of MACRS deal with the specific depreciation accounting aspects of tax law.
This section covers only the elements that materially affect after-tax economic analysis. Additional
information on how the DDB, DB, and SL methods are embedded into MACRS and how
to derive the MACRS depreciation rates is presented and illustrated in the chapter appendix,
Sections 16A.2 and 16A.3.
MACRS determines annual depreciation amounts using the relation
D t d t B [16.14]
1 R. Lundquist, “The Pedagogy of Taxes and Tax Purpose Depreciation,” Proceedings, ASEE Annual
Conference, Austin, TX, June 2009.

16.4 Modified Accelerated Cost Recovery System (MACRS) 423
where the depreciation rate d t is provided in tabulated form. As for other methods, the book value
in year t is determined by subtracting the depreciation amount from the previous year’s book
value
BV t BV t 1 D t [16.15]
or by subtracting the total depreciation from the first cost.
BV t first cost sum of accumulated depreciation
jt
B D j [16.16]
j1
MACRS has standardized and simplified many of the decisions and calculations of depreciation.
The basis B (or first cost P ) is completely depreciated; salvage is always assumed to be zero,
or S $0.
Recovery periods are standardized to specific values:
n 3, 5, 7, 10, 15, or 20 years
n 27.5 or 39 years
for personal property (e.g., equipment or vehicles)
for real property (e.g., rental property or structures)
Depreciation rates provide accelerated write-off by incorporating switching between classical
methods.
Section 16.5 explains how to determine an allowable MACRS recovery period. The MACRS
personal property depreciation rates ( d t values) for n 3, 5, 7, 10, 15, and 20 for use in Equation
[16.14] are included in Table 16–2.
MACRS depreciation rates incorporate the DDB method ( d 2 n ) and switch to SL depreciation
during the recovery period as an inherent component for personal property depreciation.
The MACRS rates start with the DDB rate or the 150% DB rate and switch when the SL method
offers faster write-off.
TABLE 16–2
Depreciation Rates d t Applied to the Basis B for the MACRS Method
Depreciation Rate (%) for Each MACRS Recovery Period in Years
Year n 3 n 5 n 7 n 10 n 15 n 20
1 33.33 20.00 14.29 10.00 5.00 3.75
2 44.45 32.00 24.49 18.00 9.50 7.22
3 14.81 19.20 17.49 14.40 8.55 6.68
4 7.41 11.52 12.49 11.52 7.70 6.18
5 11.52 8.93 9.22 6.93 5.71
6 5.76 8.92 7.37 6.23 5.29
7 8.93 6.55 5.90 4.89
8 4.46 6.55 5.90 4.52
9 6.56 5.91 4.46
10 6.55 5.90 4.46
11 3.28 5.91 4.46
12 5.90 4.46
13 5.91 4.46
14 5.90 4.46
15 5.91 4.46
16 2.95 4.46
17–20 4.46
21 2.23

424 Chapter 16 Depreciation Methods
For real property, MACRS utilizes the SL method for n 39 throughout the recovery period.
The annual percentage depreciation rate is d 139 0.02564. However, MACRS forces
partial-year recovery in years 1 and 40. The MACRS real property rates in percentage amounts are
Year 1 100 d 1 1.391%
Years 2–39 100 d t 2.564%
Year 40 100 d 40 1.177%
The real property recovery period of 27.5 years, which applies only to residential rental property,
uses the SL method in a similar fashion.
Note that all MACRS depreciation rates in Table 16–2 are presented for 1 year longer than the
stated recovery period. Also note that the extra-year rate is one-half of the previous year’s rate.
This is so because a built-in half-year convention is imposed by MACRS. This convention assumes
that all property is placed in service at the midpoint of the tax year of installation. Therefore,
only 50% of the first-year DB depreciation applies for tax purposes. This removes some of
the accelerated depreciation advantage and requires that one-half year of depreciation be taken in
year n 1.
No specially designed spreadsheet function is present for MACRS depreciation. However, the
variable declining balance (VDB) function, which is used to determine when to switch between
classical methods, can be adapted to display MACRS deprecation for each year. (The VDB function
is explained in detail in Section 16A.2 of this chapter and Appendix A of the text.) The
MACRS depreciation format of the VDB function requires embedded MAX and MIN functions,
as follows:
where
VDB( B ,0, n , MAX(0, t 1.5), MIN( n , t 0.5),d) [16.17]
B first cost
0 salvage value of S 0
n recovery period
d
2 if MACRS n 3, 5, 7, or 10
{ 1.5 if MACRS n 15 or 20
The MAX and MIN functions ensure that the MACRS half-year conventions are followed; that
is, only one-half of the first year’s depreciation is charged in year 1, and one-half of the last year’s
charge is carried over to year n 1.
EXAMPLE 16.4
Chevron Phillips Chemical Company in Baytown, Texas, acquired new equipment for its
polyethylene processing line. This chemical is a resin used in plastic pipe, retail bags, blow
molding, and injection molding. The equipment has an unadjusted basis of B $400,000, a
life of only 3 years, and a salvage value of 5% of B. The chief engineer asked the finance director
to provide an analysis of the difference between (1) the DDB method, which is the internal
book depreciation and book value method used at the plant, and (2) the required
MACRS tax depreciation and its book value. He is especially curious about the differences
after 2 years of service for this short-lived, but expensive asset. Use hand and spreadsheet
solutions to do the following:
(a) Determine which method offers the larger total depreciation after 2 years.
(b) Determine the book value for each method after 2 years and at the end of the recovery
period.
Solution by Hand
The basis is B $400,000 and the estimated S 0.05(400,000) $20,000. The MACRS rates
for n 3 are taken from Table 16–2, and the depreciation rate for DDB is d max 23
0.6667. Table 16–3 presents the depreciation and book values. Year 3 depreciation for DDB
would be $44,444(0.6667) $29,629, except this would make BV 3 $20,000. Only the
remaining amount of $24,444 is removed.

16.4 Modified Accelerated Cost Recovery System (MACRS) 425
TABLE 16–3 Comparing MACRS and DDB Depreciation, Example 16.4
Year
Rate
MACRS
Tax
Depreciation, $
Book
Value, $
DDB
Book
Depreciation, $
Book
Value, $
0 400,000 400,000
1 0.3333 133,320 266,680 266,667 133,333
2 0.4445 177,800 88,880 88,889 44,444
3 0.1481 59,240 29,640 24,444 20,000
4 0.0741 29,640 0
(a) The 2-year accumulated depreciation values from Table 16–3 are
MACRS: D 1 D 2 $133,320 177,800 $311,120
DDB: D 1 D 2 $266,667 88,889 $355,556
The DDB depreciation is larger. (Remember that for tax purposes, the company does not
have the choice in the United States of DDB as applied here.)
(b) After 2 years the book value for DDB at $44,444 is 50% of the MACRS book value of $88,880.
At the end of recovery (4 years for MACRS due to the built-in half-year convention, and
3 years for DDB), the MACRS book value is BV 4 0 and for DDB, BV 3 $20,000. This
occurs because MACRS always removes the entire first cost, regardless of the estimated salvage
value. This is a tax depreciation advantage of the MACRS method (unless the asset is
disposed of for more than the MACRS-depreciated book value, as discussed in Section 17.4).
Solution by Spreadsheet
Figure 16–4 presents the spreadsheet solution using the VDB function (column B) for MACRS
depreciation (in lieu of the MACRS rates) and applying the DDB function in column D.
(a) The 2-year accumulated depreciation values are
MACRS (add cells B6 B7): $133,333 177,778 $311,111
DDB (add cells D6 D7): $266,667 88,889 $355,556
(b) Book values after 2 years are
MACRS (cell C7): $88,889
DDB (cell E7): $44,444
The book values are plotted in Figure 16–4. Observe that MACRS goes to zero in year 4, while
DDB stops at $20,000 in year 3.
Comment
It is advisable to set up a spreadsheet template for use with depreciation problems in this and
future chapters. The format and functions of Figure 16–4 are a good template for MACRS and
DDB methods.
Figure 16–4
Spreadsheet screen shot of MACRS and DDB depreciation and book value, Example 16.4.

426 Chapter 16 Depreciation Methods
MACRS simplifies depreciation computations, but it removes much of the flexibility of
method selection for a business or corporation. In general, an economic comparison that includes
depreciation may be performed more rapidly and usually without altering the final decision by
applying the classical straight line method in lieu of MACRS.
16.5 Determining the MACRS Recovery Period
The expected useful life of property is estimated in years and used as the n value in alternative
evaluation and in depreciation computations. For book depreciation the n value should be the
expected useful life. However, when the depreciation will be claimed as tax deductible, the
n value should be lower. There are tables that assist in determining the life and recovery period
for tax purposes.
The advantage of a recovery period shorter than the anticipated useful life is leveraged by the
accelerated depreciation methods that write off more of the basis B in the initial years.
The U.S. government requires that all depreciable property be classified into a property class
which identifies its MACRS-allowed recovery period. Table 16–4, a summary of material from
IRS Publication 946, gives examples of assets and the MACRS n values. Virtually any property
considered in an economic analysis has a MACRS n value of 3, 5, 7, 10, 15, or 20 years.
Table 16–4 provides two MACRS n values for each property. The first is the general depreciation
system (GDS) value, which we use in examples and problems. The depreciation rates in Table
16–2 correspond to the n values for the GDS column and provide the fastest write-off allowed. The
rates utilize the DDB method or the 150% DB method with a switch to SL depreciation. Note that
any asset not in a stated class is automatically assigned a 7-year recovery period under GDS.
The far right column of Table 16–4 lists the alternative depreciation system ( ADS ) recovery
period range. This alternative method allows the use of SL depreciation over a longer
TABLE 16–4
Example MACRS Recovery Periods for Various Asset Descriptions
MACRS
n Value, Years
Asset Description (Personal and Real Property) GDS ADS Range
Special manufacturing and handling devices, tractors,
3 3–5
racehorses
Computers and peripherals, oil and gas drilling
5 6–9.5
equipment, construction assets, autos, trucks, buses,
cargo containers, some manufacturing equipment
Office furniture; some manufacturing equipment;
7 10–15
railroad cars, engines, tracks; agricultural machinery;
petroleum and natural gas equipment; all
property not in another class
Equipment for water transportation, petroleum refining,
10 15–19
agriculture product processing, durable-goods
manufacturing, shipbuilding
Land improvements, docks, roads, drainage, bridges,
15 20–24
landscaping, pipelines, nuclear power production
equipment, telephone distribution
Municipal sewers, farm buildings, telephone switching
20 25–50
buildings, power production equipment (steam
and hydraulic), water utilities
Residential rental property (house, mobile home) 27.5 40
Nonresidential real property attached to the land, but
not the land itself
39 40

16.6 Depletion Methods 427
recovery period than the GDS. The half-year convention applies, and any salvage value is neglected,
as it is in regular MACRS. The use of ADS is generally a choice left to a company, but it is required
for some special asset situations. Since it takes longer to depreciate the asset, and since the SL model
is required (thus removing the advantage of accelerated depreciation), ADS is usually not considered
an option for the economic analysis. This electable SL option is, however, sometimes chosen
by businesses that are young and do not need the tax benefit of accelerated depreciation during the
first years of operation and asset ownership. If ADS is selected, tables of d t rates are available.
16.6 Depletion Methods
Previously, for all assets, facilities, and equipment that can be replaced we have applied depreciation.
We now turn to irreplaceable natural resources and the equivalent of depreciation, which
is called depletion.
Depletion is a book method (noncash) to represent the decreasing value of a natural resource
as it is recovered, removed, or felled. The two methods of depletion for book or tax purposes are
used to write off the first cost, or value of the estimated quantity, of resources in mines, wells,
quarries, geothermal deposits, forests, and the like.
The two methods of depletion are cost and percentage depletion , as described below. Details for
U.S. taxes on depletion are found in IRS Publication 535, Business Expenses .
Cost depletion Sometimes referred to as factor depletion, cost depletion is based on the level
of activity or usage, not time, as in depreciation. Cost depletion may be applied to most types of
natural resources and must be applied to timber production. The cost depletion factor for year t , denoted
by CD t , is the ratio of the first cost of the resource to the estimated number of units recoverable.
CD t ————————
first cost
resource capacity
[16.18]
The annual depletion charge is CD t times the year’s usage or volume. The total cost depletion
cannot exceed the fi rst cost of the resource. If the capacity of the property is reestimated some
year in the future, a new cost depletion factor is determined based upon the undepleted amount
and the new capacity estimate.
Percentage depletion This is a special consideration given for natural resources. A constant,
stated percentage of the resource’s gross income may be depleted each year provided it does not
exceed 50% of the company’s taxable income. The depletion amount for year t is calculated as
Percentage depletion t percentage depletion rate
gross income from property
PD GI t [16.19]
Using percentage depletion, total depletion charges may exceed first cost with no limitation. The
U.S. government does not generally allow percentage depletion to be applied to oil and gas wells
(except small independent producers).
The annual percentage depletion rates for some common natural deposits are listed below per
U.S. tax law.
Deposit
Percentage of
Gross Income, PD
Sulfur, uranium, lead, nickel, zinc,
22
and some other ores and minerals
Gold, silver, copper, iron ore, and
15
some oil shale
Oil and natural gas wells (varies) 15–22
Coal, lignite, sodium chloride 10
Gravel, sand, peat, some stones 5
Most other minerals, metallic ores 14

428 Chapter 16 Depreciation Methods
EXAMPLE 16.5
Temple-Inland Corporation has negotiated the rights to cut timber on privately held forest acreage
for $700,000. An estimated 350 million board feet of lumber is harvestable.
(a) Determine the depletion amount for the first 2 years if 15 million and 22 million board feet
are removed.
(b) After 2 years the total recoverable board feet was reestimated upward to be 450 million from
the time the rights were purchased. Compute the new cost depletion factor for years 3 and later.
Solution
(a) Use Equation [16.18] for CD t in dollars per million board feet.
700,000
CD t ———— $2000 per million board feet
350
Multiply CD t by the annual harvest to obtain depletion of $30,000 in year 1 and $44,000
in year 2. Continue until a total of $700,000 is written off.
(b) After 2 years, a total of $74,000 has been depleted. A new CD t value must be calculated
based on the remaining 700,000 74,000 $626,000 investment. Additionally, with the
new estimate of 450 million board feet, a total of 450 15 22 413 million board feet
remains. For years t 3, 4, . . . , the cost depletion factor is
CD t ————
626,000 $1516 per million board feet
413
EXAMPLE 16.6
A gold mine was purchased for $10 million. It has an anticipated gross income of $5.0 million
per year for years 1 to 5 and $3.0 million per year after year 5. Assume that depletion charges
do not exceed 50% of taxable income. Compute annual depletion amounts for the mine. How
long will it take to recover the initial investment at i 0%?
Solution
The rate for gold is PD 0.15. Depletion amounts are
Years 1 to 5: 0.15(5.0 million) $750,000
Years thereafter: 0.15(3.0 million) $450,000
A total of $3.75 million is written off in 5 years, and the remaining $6.25 million is written off
at $450,000 per year. The total number of years is
$6.25 million
5 —————— 5 13.9 18.9
$450,000
In 19 years, the initial investment could be fully depleted.
In many of the natural resource depletion situations, the tax law allows the larger of the two
depletion amounts to be claimed each year. This is allowed provided the percentage depletion
amount does not exceed 50% of taxable income. Therefore, it is wise to calculate both depletion
amounts and select the larger. Use the following terminology for year t ( t 1, 2, . . .).
CDA t cost depletion amount
PDA t percentage depletion amount
TI t taxable income
The guideline for the tax-allowed depletion amount for year t is
Depletion { max[CDA t , PDA t ]
max[CDA t , 50% of TI t ]
if PDA t 50% of TI t
if PDA t 50% of TI t

For example, assume a medium-sized quarry owner calculates the following for 1 year.
TI $500,000 CDA $275,000 PDA $280,000
Since 50% of TI is $250,000, the PDA is too large and, therefore, is not allowed. For tax purposes,
apply the guideline above and use the cost depletion of $275,000, since it is larger than
50% of TI.
Chapter Summary 429
CHAPTER SUMMARY
Depreciation may be determined for internal company records (book depreciation) or for income
tax purposes (tax depreciation). In the United States, the MACRS method is the only
one allowed for tax depreciation. In many other countries, straight line and declining balance
methods are applied for both tax and book depreciation. Depreciation does not result in cash
flow directly. It is a book method by which the capital investment in tangible property is recovered.
The annual depreciation amount is tax deductible, which can result in actual cash
flow changes.
Some important points about the straight line, declining balance, and MACRS methods are
presented below. Common relations for each method are summarized in Table 16–5.
Straight Line (SL)
• It writes off capital investment linearly over n years.
• The estimated salvage value is always considered.
• This is the classical, nonaccelerated depreciation model.
Declining Balance (DB)
• The method accelerates depreciation compared to the straight line method.
• The book value is reduced each year by a fixed percentage.
• The most used rate is twice the SL rate, which is called double declining balance (DDB).
• It has an implied salvage that may be lower than the estimated salvage.
• It is not an approved tax depreciation method in the United States. It is frequently used for
book depreciation purposes.
Modified Accelerated Cost Recovery System (MACRS)
• It is the only approved tax depreciation system in the United States.
• It automatically switches from DDB or DB to SL depreciation.
• It always depreciates to zero; that is, it assumes S 0.
• Recovery periods are specified by property classes.
• Depreciation rates are tabulated.
• The actual recovery period is 1 year longer due to the imposed half-year convention.
• MACRS straight line depreciation is an option, but recovery periods are longer than those for
regular MACRS.
Cost and percentage depletion methods recover investment in natural resources. The annual
cost depletion factor is applied to the amount of resource removed. No more than the initial investment
can be recovered with cost depletion. Percentage depletion, which can recover more
than the initial investment, reduces the investment value by a constant percentage of gross income
each year.
TABLE 16–5
Summary of Common Depreciation Method Relations
Method MACRS SL DDB
Fixed depreciation rate d Not defined — 1 2
n —
n
Annual rate d t Table 16–2 —
n
1
Annual depreciation D t d tB ——— B n
S
d(1 d)t1
d(BVt1)
Book value BV t BV t1 D t B tD t B(1 d) t

430 Chapter 16 Depreciation Methods
CHAPTER 16 APPENDIX
16A.1 Sum-of-Years-Digits (SYD) and Unit-of-
Production (UOP) Depreciation
The SYD method is a historical accelerated depreciation technique that removes much of the
basis in the first one-third of the recovery period; however, write-off is not as rapid as for DDB
or MACRS. This technique may be used in an engineering economy analysis in the book depreciation
of multiple-asset accounts (group and composite depreciation).
The mechanics of the method involve the sum of the year’s digits from 1 through the recovery
period n . The depreciation charge for any given year is obtained by multiplying the basis of the
asset, less any salvage value, by the ratio of the number of years remaining in the recovery period
to the sum of the year’s digits SUM.
depreciable years remaining
D t ————————————— (basis salvage value)
sum of years digits
D t ————— n t 1 (B S) [16A.1]
SUM
where SUM is the sum of the digits 1 through n .
The book value for any year t is calculated as
BV t B
jn
n(n 1)
SUM j ————
2
j1
t(n t2 0.5)
——————— (B S) [16A.2]
SUM
The rate of depreciation decreases each year and equals the multiplier in Equation [16A.1].
d t ———— n t 1
[16A.3]
SUM
The SYD spreadsheet function displays the depreciation for the year t . The function format is
SYD( B , S , n , t )
EXAMPLE 16A.1
Calculate the SYD depreciation charges for year 2 for electro-optics equipment with
B $25,000, S $4000, and an 8-year recovery period.
Solution
The sum of the year’s digits is 36, and the depreciation amount for the second year by Equation
[16A.1] is
D 2 —— 7 (21,000) $4083
36
The SYD function is SYD(25000,4000,8,2).
Figure 16A–1 is a plot of the book values for an $80,000 asset with S $10,000 and n
10 years using the four depreciation methods that we have learned. The MACRS, DDB, and SYD
curves track closely except for year 1 and years 9 through 11.

16A.1 Sum-of-Years-Digits (SYD) and Unit-of-Production (UOP) Depreciation 431
80
70
MACRS
Book value BV t $1000
60
50
40
30
20
DDB
SYD
SL
10
S = $10,000
0 1 2 3 4 5 6 7 8 9 10
Year
Figure 16A–1
Comparison of book values using SL, SYD, DDB, and MACRS depreciation.
11
A second depreciation method that is not allowed for tax purposes, but useful in some situations
is the unit-of-production (UOP) method . When the decreasing value of equipment is
based on usage, not time , the UOP method is quite applicable. Suppose a highway contractor
has a series of state highway department contracts that will last several years and that earth moving
equipment is purchased for use on all contracts. If the equipment usage goes up and down
significantly over the years, the UOP method is ideal for book depreciation. For year t , UOP
deprecation is calculated as
actual usage for year t
D t —————————— (basis salvage)
total lifetime usage
[16A.4]
EXAMPLE 16A.2
Zachry Contractors purchased an $80,000 mixer for use during the next 10 years for contract
work on IH-10 in San Antonio. The mixer will have a negligible salvage value after 10 years,
and the total amount of material to process is estimated at 2 million m 3 . Use the actual usage per
year shown in Table 16A–1 and the unit-of-production method to determine annual depreciation.
Solution
The actual usage each year is placed in the numerator of Equation [16A.4] to determine the annual
depreciation based on the estimated total lifetime amount of material, 2 million m 3 in this
case. Table 16A–1 shows the annual and cumulative depreciation over the 10 years. If the mixer
is continued in service after the 2 million m 3 is processed, no further depreciation is allowed.
TABLE 16A–1 Unit-of-Production Method of Depreciation, Example 16A.2
Year t
Actual
Usage, 1000 m 3
Annual
Depreciation D t , $
Cumulative
Depreciation, $
1 400 16,000 16,000
2–8 200 8,000 72,000
9–10 100 4,000 80,000
Total 2000 80,000

432 Chapter 16 Depreciation Methods
16A.2 Switching between Depreciation Methods
Switching between depreciation methods may assist in accelerated reduction of the book value.
It also maximizes the present value of accumulated and total depreciation over the recovery period.
Therefore, switching usually increases the tax advantage in years where the depreciation is
larger. The approach below is an inherent part of MACRS.
Switching from a DB method to the SL method is the most common switch because it usually
offers a real advantage, especially if the DB method is DDB. General rules of switching are summarized
here.
1. Switching is recommended when the depreciation for year t by the currently used method is
less than that for a new method. The selected depreciation D t is the larger amount.
2. Only one switch can take place during the recovery period.
3. Regardless of the (classical) depreciation methods, the book value cannot go below the
estimated salvage value. When switching from a DB method, the estimated salvage value,
not the DB-implied salvage value, is used to compute the depreciation for the new method;
we assume S 0 in all cases. (This does not apply to MACRS, since it already includes
switching.)
4. The undepreciated amount, that is, BV t , is used as the new adjusted basis to select the larger
D t for the next switching decision.
In all situations, the criterion is to maximize the present worth of the total depreciation PW D .
The combination of depreciation methods that results in the largest present worth is the best
switching strategy.
tn
PW D D t (PF, i, t)
t1
[16A.5]
This logic minimizes tax liability in the early part of an asset’s recovery period.
Switching is most advantageous from a rapid write-off method such as DDB to the SL model.
This switch is predictably advantageous if the implied salvage value computed by Equation
[16.11] exceeds the salvage value estimated at purchase time; that is, switch if
BV n B (1 d ) n estimated S
[16A.6]
Since we assume that S will be zero per rule 3 above, and since BV n will be greater than zero, for
a DB method a switch to SL is always advantageous. Depending upon the values of d and n , the
switch may be best in the later years or last year of the recovery period, which removes the implied
S inherent to the DDB model.
The procedure to switch from DDB to SL depreciation is as follows:
1. For each year t , compute the two depreciation charges.
For DDB: D DDB d (BV t 1) [16A.7]
For SL:
BV t1
D SL ————
n t 1
2. Select the larger depreciation value. The depreciation for each year is
D t max[ D DDB , D SL ]
[16A.8]
[16A.9]
3. If needed, determine the present worth of total depreciation, using Equation [16A.5].
It is acceptable, though not usually financially advantageous, to state that a switch will take
place in a particular year, for example, a mandated switch from DDB to SL in year 7 of a 10-year
recovery period. This approach is usually not taken, but the switching technique will work
correctly for all depreciation methods.
To use a spreadsheet for switching, first understand the depreciation model switching rules
and practice the switching procedure from declining balance to straight line. Once these are understood,
the mechanics of the switching can be speeded up by applying the spreadsheet function

16A.2 Switching between Depreciation Methods 433
VDB (variable declining balance). This is a quite powerful function that determines the depreciation
for 1 year or the total over several years for the DB-to-SL switch. The function format is
VDB(B, S, n,start_t,end_t, d,no_switch)
[16A.10]
Appendix A explains all the fields in detail, but for simple applications, where the DDB and SL
annual D t values are needed, the following are correct entries:
start_t is the year (t1)
end_t is year t
d is optional; 2 for DDB is assumed, the same as in the DDB function
no_switch is an optional logical value:
FALSE or omitted—switch to SL occurs, if advantageous
TRUE—DDB or DB method is applied with no switching to SL depreciation considered.
Entering TRUE for the no_switch option obviously causes the VDB function to display the same
depreciation amounts as the DDB function. This is discussed in Example 16A.3 d . You may notice
that the VDB function is the same one used to calculate annual MACRS depreciation.
EXAMPLE 16A.3
The Outback Steakhouse main office has purchased a $100,000 online document imaging system
with an estimated useful life of 8 years and a tax depreciation recovery period of 5 years.
Compare the present worth of total depreciation for (a) the SL method, (b) the DDB method,
and (c) DDB-to-SL switching. (d) Perform the DDB-to-SL switch using a spreadsheet and plot
the book values. Use a rate of i 15% per year.
Solution by Hand
The MACRS method is not involved in this solution.
(a) Equation [16.1] determines the annual SL depreciation.
D t ——————
100,000 0 $20,000
5
Since D t is the same for all years, the PA factor replaces PF to compute PW D .
PW D 20,000(PA,15%,5) 20,000(3.3522) $67,044
(b) For DDB, d 25 0.40. The results are shown in Table 16A–2. The value PW D
$69,915 exceeds $67,044 for SL depreciation. As is predictable, the accelerated depreciation
of DDB increases PW D.
(c) Use the DDB-to-SL switching procedure.
1. The DDB values for D t in Table 16A–2 are repeated in Table 16A–3 for comparison
with the D SL values from Equation [16A.8]. The D SL values change each year because
BV t1 is different. Only in year 1 is D SL $20,000, the same as computed in part (a).
For illustration, compute D SL values for years 2 and 4. For t 2, BV 1 $60,000 by the
DDB method and
D SL —————
60,000 0
5 2 1 $15,000
For t 4, BV 3 $21,600 by the DDB method and
D SL —————
21,600 0
5 4 1 $10,800
2. The column “Larger D t” indicates a switch in year 4 with D 4 $10,800. The D SL
$12,960 in year 5 would apply only if the switch occurred in year 5. Total depreciation
with switching is $100,000 compared to the DDB amount of $92,224.
3. With switching, PW D $73,943, which is an increase over both the SL and DDB
methods.

434 Chapter 16 Depreciation Methods
TABLE 16A–2
DDB Model Depreciation and Present Worth Computations,
Example 16A.3b
Year
t D t , $ BV t , $ (PF,15%,t)
Present Worth
of D t , $
0 100,000
1 40,000 60,000 0.8696 34,784
2 24,000 36,000 0.7561 18,146
3 14,400 21,600 0.6575 9,468
4 8,640 12,960 0.5718 4,940
5 5,184 7,776 0.4972 2,577
Totals 92,224 69,915
TABLE 16A–3
Depreciation and Present Worth for DDB-to-SL Switching,
Example 16A.3c
Year
t
DDB Method, $
BV t
D DDB
SL Method
D SL, $
Larger
D t , $
PF
Factor
Present
Worth of
D t, $
0 — 100,000
1 40,000 60,000 20,000 40,000 0.8696 34,784
2 24,000 36,000 15,000 24,000 0.7561 18,146
3 14,400 21,600 12,000 14,400 0.6575 9,468
4* 8,640 12,960 10,800 10,800 0.5718 6,175
5 5,184 7,776 12,960 10,800 0.4972 5,370
Totals 92,224 100,000 73,943
*Indicates year of switch from DDB to SL depreciation.
Solution by Spreadsheet
(d) In Figure 16A–2, column D entries are the VDB functions to determine that the DDB-to-
SL switch should take place in year 4. The entries “2,FALSE” at the end of the VDB function
are optional (see the VDB function description). If TRUE were entered, the declining
balance model would be maintained throughout the recovery period, and the annual depreciation
amounts would be equal to those in column B. The plot in Figure 16A–2 indicates
another difference in depreciation methods. The terminal book value in year 5 for the DDB
method is BV 5 $7776, while the DDB-to-SL switch reduces the book value to zero.
The NPV function determines the PW of depreciation (row 9). The results here are the
same as in parts (b) and (c) above. The DDB-to-SL switch has the larger PW D value.
DDB($C$3,0,5,$A7)
VDB($E$3,0,5,$A6,$A7,2,FALSE)
Figure 16A–2
Depreciation for DDB-to-SL switch using the VDB function, Example 16A.3.

16A.3 Determination of MACRS Rates 435
In MACRS, recovery periods of 3, 5, 7, and 10 years apply DDB depreciation with half-year
convention switching to SL. When the switch to SL takes place, which is usually in the last 1 to
3 years of the recovery period, any remaining basis is charged off in year n 1 so that the book
value reaches zero. Usually 50% of the applicable SL amount remains after the switch has occurred.
For recovery periods of 15 and 20 years, 150% DB with the half-year convention and the
switch to SL apply.
The present worth of depreciation PW D will always indicate which method is the most advantageous.
Only the MACRS rates for the GDS recovery periods (Table 16–4) utilize the DDBto-SL
switch. The MACRS rates for the alternative depreciation system (ADS) have longer
recovery periods and impose the SL model for the entire recovery period.
EXAMPLE 16A.4
In Example 16A.3, parts (c) and (d), the DDB-to-SL switching method was applied to a
$100,000, n 5 years asset resulting in PW D $73,943 at i 15%. Use MACRS to depreciate
the same asset for a 5-year recovery period, and compare PW D values.
Solution
Table 16A–4 summarizes the computations for depreciation (using Table 16–2 rates), book
value, and present worth of depreciation. The PW D values for all four methods are as follows:
DDB-to-SL switching $73,943
Double declining balance $69,916
MACRS $69,016
Straight line $67,044
MACRS provides a slightly less accelerated write-off. This is so, in part, because the half-year
convention disallows 50% of the first-year DDB depreciation (which amounts to 20% of the
basis). Also the MACRS recovery period extends to year 6, further reducing PW D .
TABLE 16A–4
Depreciation and Book Value Using MACRS,
Example 16A.4
t d t D t , $ BV t, $
0 — — 100,000
1 0.20 20,000 80,000
2 0.32 32,000 48,000
3 0.192 19,200 28,800
4 0.1152 11,520 17,280
5 0.1152 11,520 5,760
6 0.0576 5,760 0
1.000 100,000
t6
PW D D t(PF,15%,t) $69,016
t1
16A.3 Determination of MACRS Rates
The depreciation rates for MACRS incorporate the DB-to-SL switching for all GDS recovery
periods from 3 to 20 years. In the first year, some adjustments have been made to compute the
MACRS rate. The adjustments vary and are not usually considered in detail in economic analyses.
The half-year convention is always imposed, and any remaining book value in year n is removed
in year n 1. The value S 0 is assumed for all MACRS schedules.
Since different DB depreciation rates apply for different n values, the following summary may
be used to determine D t and BV t values. The symbols D DB and D SL are used to identify DB and
SL depreciation, respectively.

436 Chapter 16 Depreciation Methods
For n 3, 5, 7, and 10 Use DDB depreciation with the half-year convention, switching to SL
depreciation in year t when D SL D DB . Use the switching rules of Section 16A.2, and add onehalf
year when computing D SL to account for the half-year convention. The yearly depreciation
rates are
{
— 1 n t 1
d t
2—
n t 2, 3, . . .
[16A.11]
Annual depreciation values for each year t applied to the adjusted basis, allowing for the halfyear
convention, are
D DB d t (BV t 1 )
{
— 1
2 (
— 1 n ) B t 1
D SL
BV t1
————— t 2, 3, . . . , n
n t 1.5
[16A.12]
[16A.13]
After the switch to SL depreciation takes place—usually in the last 1 to 3 years of the recovery
period—any remaining book value in year n is removed in year n 1.
For n 15 and 20 Use 150% DB with the half-year convention and the switch to SL when
D SL D DB . Until SL depreciation is more advantageous, the annual DB depreciation is computed
using a form of Equation [16A.7]
D DB d t (BV t 1 )
where
{
—— 0.75 n t 1
d t
—— 1.50
n t 2, 3, . . .
[16A.14]
EXAMPLE 16A.5
A wireless tracking system for shop floor control with a MACRS 5-year recovery period has
been purchased for $10,000. (a) Use Equations [16A.11] through [16A.13] to obtain the annual
depreciation and book value. (b) Determine the resulting annual depreciation rates and compare
them with the MACRS rates in Table 16–2 for n 5.
Solution
(a) With n 5 and the half-year convention, use the DDB-to-SL switching procedure to obtain
the results in Table 16A–5. The switch to SL depreciation, which occurs in year 4
when both depreciation values are equal, is indicated by
D DB 0.4(2880) $1152
D SL ————— 2880
5 4 1.5 $1152
The SL depreciation of $1000 in year 1 results from applying the half-year convention included
in the first relation of Equation [16A.13]. Also, the SL depreciation of $576 in
year 6 is the result of the half-year convention.
(b) The actual rates are computed by dividing the “Larger D t ” column values by the first cost
of $10,000. The rates below are the same as the Table 16–2 rates.
t 1 2 3 4 5 6
d t 0.20 0.32 0.192 0.1152 0.1152 0.0576

16A.3 Determination of MACRS Rates 437
TABLE 16A–5
Depreciation Amounts Used to Determine MACRS Rates
for n 5, Example 16A.5
Years
DDB
SL Depreciation Larger
t d t D DB, $ D SL, $
D t, $ BV t, $
0 — — — — 10,000
1 0.2 2,000 1,000 2,000 8,000
2 0.4 3,200 1,777 3,200 4,800
3 0.4 1,920 1,371 1,920 2,880
4 0.4 1,152 1,152 1,152 1,728
5 0.4 691 1,152 1,152 576
6 — — 576 576 0
10,000
It is clearly easier to use the rates in Table 16–2 or the VDB spreadsheet function than to determine
each MACRS rate using the switching logic above. But the logic behind the MACRS
rates is described here for those interested. The annual MACRS rates may be derived by using
the applicable rate for the DB method. The subscripts DB and SL have been inserted along with
the year t . For the first year t 1,
d DB,1 1 —
n
or d SL,1 1 —
2 ( 1 —
n )
For summation purposes only, we introduce the subscript i ( i 1, 2, . . . , t ) on d . Then the depreciation
rates for years t 2, 3, . . . , n are
it1
d DB,t d ( 1
i1
it1
( 1
i1
d SL,t
d i
)
d i
)
—————
n t 1.5
Also, for year n 1, the MACRS rate is one-half the SL rate of the previous year n.
[16A.15]
[16A.16]
d SL,n1 0.5(d SL,n )
[16A.17]
The DB and SL rates are compared each year to determine which is larger and when the switch
to SL depreciation should occur.
EXAMPLE 16A.6
Verify the MACRS rates in Table 16–2 for a 3-year recovery period. The rates in percent are
33.33, 44.45, 14.81, and 7.41.
Solution
The fixed rate for DDB with n 3 is d 23 0.6667. Using the half-year convention in year 1
and Equations [16A.15] through [16A.17], the results are as follows:
d 1 : d DB,1 0.5d 0.5(0.6667) 0.3333
d 2 : Cumulative depreciation rate is 0.3333.
d DB,2 0.6667(1 0.3333) 0.4445
d SL,2 ————— 1 0.3333
3 2 1.5 0.2267
(larger value)

438 Chapter 16 Depreciation Methods
d 3 : Cumulative depreciation rate is 0.3333 0.4445 0.7778.
d DB,3 0.6667(1 0.7778) 0.1481
d SL,2 1 0.7778 —————
3 3 1.5 0.1481
Both values are the same; switch to straight line depreciation.
d 4: This rate is 50% of the last SL rate.
d 4 0.5(d SL,3) 0.5(0.1481) 0.0741
PROBLEMS
Fundamentals of Depreciation
16.1 How does depreciation affect a company’s cash
flow?
16.2 What is the difference between book value and
market value?
16.3 State the difference between book depreciation
and tax depreciation.
16.4 State the difference between unadjusted and adjusted
basis.
16.5 Explain why the recovery period used for tax depreciation
purposes may be different from the estimated
n value in an engineering economy study.
16.6 Visit the U.S. Internal Revenue Service website at
www.irs.gov and answer the following questions
about depreciation and MACRS by consulting
Publication 946, How to Depreciate Property .
(a) What is the definition of depreciation according
to the IRS?
(b) What is the description of the term salvage
value ?
(c) What are the two depreciation systems within
MACRS, and what are the major differences
between them?
(d) What are the properties listed that cannot be
depreciated under MACRS?
(e) When does depreciation begin and end?
(f) What is a Section 179 deduction?
16.7 A major energy production company has the following
information regarding the acquisition of
new-generation equipment.
Purchase price $580,000
Transoceanic shipping and delivery cost $4300
Installation cost (1 technician at $1600 per day
for 4 days) $6400
Tax recovery period 15 years
Book depreciation recovery period 10 years
Salvage value 10% of purchase price
Operating cost (with technician) $185,000
per year
The manager of the department asked the newest
hire to enter the appropriate data in the tax accounting
program. For the MACRS method, what
are the values of B , n , and S in depreciating the
asset for tax purposes?
16.8 Stahmann Products paid $350,000 for a numerical
controller during the last month of 2007 and had it
installed at a cost of $50,000. The recovery period
was 7 years with an estimated salvage value of
10% of the original purchase price. Stahmann sold
the system at the end of 2011 for $45,000.
(a) What numerical values are needed to develop
a depreciation schedule at purchase time?
(b) State the numerical values for the following:
remaining life at sale time, market value in
2011, book value at sale time if 65% of the
basis had been depreciated.
16.9 An asset with an unadjusted basis of $50,000 was
depreciated over n tax 10 years for tax depreciation
purposes and n book 5 years for book depreciation
purposes. The annual depreciation was
1 n using the relevant life value. Use a spreadsheet
to plot on one graph the annual book value
for both methods of depreciation.
Straight Line Depreciation
16.10 What is the depreciation rate d t per year for an
asset that has an 8-year useful life and is straight
line depreciated?
16.11 Pneumatics Engineering purchased a machine that
had a first cost of $40,000, an expected useful life
of 8 years, a recovery period of 10 years, and a
salvage value of $10,000. The operating cost of the
machine is expected to be $15,000 per year. The
inflation rate is 6% per year and the company’s

Problems 439
MARR is 11% per year. Determine ( a ) the depreciation
charge for year 3, ( b ) the present worth of
the third-year depreciation charge in year 0, the
time of asset purchase, and ( c ) the book value for
year 3 according to the straight line method.
16.12 An asset that is book-depreciated over a 5-year period
by the straight line method has BV 3 $62,000
with a depreciation charge of $26,000 per year.
Determine ( a ) the first cost of the asset and ( b ) the
assumed salvage value.
16.13 Lee Company of Westbrook, Connecticut, manufactures
pressure relief inserts for thermal relief
and low-flow hydraulic pressure relief applications
where zero leakage is required. A machine purchased
3 years ago has been book-depreciated by
the straight line method using a 5-year useful life.
If the book value at the end of year 3 is $30,000 and
the company assumed that the machine would be
worthless at the end of its 5-year useful life,
( a ) what is the book depreciation charge each year
and ( b ) what was the first cost of the machine?
16.14 An asset has an unadjusted basis of $200,000, a
salvage value of $10,000, and a recovery period of
7 years. Write a single-cell spreadsheet function to
display the book value after 5 years of straight line
depreciation. Use your function to determine the
book value.
16.15 Bristol Myers Squibb purchased a tablet-forming
machine in 2008 for $750,000. The company
planned to use the machine for 10 years; however,
due to rapid obsolescence it will be retired after
only 4 years in 2012. Develop a spreadsheet for
depreciation and book value amounts necessary to
answer the following.
(a) What is the amount of capital investment remaining
when the asset is prematurely retired?
(b) If the asset is sold at the end of 4 years for
$175,000, what is the amount of capital investment
lost based on straight line depreciation?
( c) If the new-technology machine has an estimated
cost of $300,000, how many more
years should the company retain and depreciate
the currently owned machine to make its
book value and the first cost of the new machine
equal to each other?
16.16 A special-purpose graphics workstation acquired
by Busbee Consultants has B $50,000 with a
4-year recovery period. Tabulate the values for SL
depreciation, accumulated depreciation, and book
value for each year if ( a ) S 0 and ( b ) S $16,000.
( c ) Use a spreadsheet to plot the book value over
the 4 years on one chart for both salvage value
estimates.
16.17 A company owns the same asset in a U.S. plant
(New York) and in a EU plant (Paris). It has B
$2,000,000 and a salvage value of 20% of B . For
tax depreciation purposes, the United States allows
a straight line write-off over 5 years, while
the EU allows SL write-off over 8 years. The general
managers of the two plants want to know the
difference in ( a ) the depreciation amount for year
5 and ( b ) the book value after 5 years. Using a
spreadsheet, write cell functions in only two cells
to answer both questions.
Declining Balance Depreciation
16.18 When declining balance (DB) depreciation is applied,
there can be three different depreciation
rates involved— d , d max , and d t . Explain the differences
between these rates.
16.19 Equipment for immersion cooling of electronic
components has an installed value of $182,000
with an estimated trade-in value of $40,000 after
15 years. For years 2 and 10, use DDB book depreciation
to determine ( a ) the depreciation charge
and ( b ) the book value.
16.20 A cooling-water pumping station at the LCRA
plant costs $600,000 to construct, and it is projected
to have a 25-year life with an estimated
salvage value of 15% of the construction cost.
However, the station will be book-depreciated to
zero over a recovery period of 30 years. Calculate
the annual depreciation charge for years 4, 10,
and 25, using ( a ) straight line depreciation and
( b ) DDB depreciation. ( c ) What is the implied
salvage value for DDB? ( d ) Use a spreadsheet to
build the depreciation and book value schedules
for both methods to verify your answers.
16.21 A video recording system was purchased 3 years
ago at a cost of $30,000. A 5-year recovery period
and DDB depreciation have been used to write off
the basis. The system is to be replaced this year with
a trade-in value of $5000. What is the difference between
the book value and the trade-in value?
16.22 An engineer with Accenture Middle East BV in
Dubai was asked by her client to help him understand
the difference between 150% DB and DDB depreciation.
Answer these questions if B $180,000,
n 12 years, and S $30,000.
( a) What are the book values after 12 years for
both methods?
( b) How do the estimated salvage and these book
values compare in value after 12 years?
( c) Which of the two methods, when calculated
correctly considering S $30,000, writes off
more of the first cost over 12 years?

440 Chapter 16 Depreciation Methods
16.23 Exactly 10 years ago, Boyditch Professional Associates
purchased $100,000 in depreciable assets
with an estimated salvage of $10,000. For tax
depreciation, the SL method with n 10 years
was used, but for book depreciation, Boyditch applied
the DDB method with n 7 years and neglected
the salvage estimate. The company sold
the assets today for $12,500.
(a) Compare the sales price today with the book
values using the SL and DDB methods.
(b) If the salvage of $12,500 had been estimated
exactly 10 years ago, determine the depreciation
for each method in year 10.
16.24 Shirley is studying depreciation in her engineering
management course. The instructor asked her to
graphically compare the total percent of first cost
depreciated for an asset costing B dollars over a
life of n 5 years for DDB and 125% DB depreciation.
Help her by developing the plots of percent
of B depreciated versus years. Use a spreadsheet
unless otherwise instructed.
MACRS Depreciation
16.25 Explain the difference between an accelerated depreciation
method and one that is not accelerated.
Give an example of each.
16.26 What was one of the prime reasons that MACRS
depreciation was initiated in the mid-1980s?
16.27 A company just purchased an intelligent robot,
which has a first cost of $80,000. Since the robot is
unique in its capabilities, the company expects to
be able to sell it in 4 years for $95,000.
(a) If the company spends $10,000 per year in
maintenance and operation of the robot, what
will the company’s MACRS depreciation
charge be in year 2? Assume the recovery period
for robots is 5 years and the company’s
MARR is 16% per year when the inflation
rate is 9% per year.
(b) Determine the book value of the robot at the
end of year 2.
16.28 Animatics Corp. of Santa Clara, California, makes
small servo systems with built-in controllers, amplifiers,
and encoders so that they can control entire
machines. The company purchased an asset 2 years
ago that has a 5-year recovery period. The depreciation
charge by the MACRS method for year 2 is $24,320.
(a) What was the first cost of the asset?
(b) How much was the depreciation charge in
year 1?
(c) Develop the complete MACRS depreciation
and book value schedule using the VDB
function.
16.29 A plant manager for a large cable company knows
that the remaining invested value of certain types
of manufacturing equipment is more closely approximated
when the equipment is depreciated
linearly by the SL method compared to a rapid
write-off method such as MACRS. Therefore, he
keeps two sets of books, one for tax purposes
(MACRS) and one for equipment management
purposes (SL). For an asset that has a first cost of
$80,000, a depreciable life of 5 years, and a salvage
value equal to 25% of the first cost, determine
the difference in the book values shown in
the two sets of books at the end of year 4.
16.30 The manager of a Glidden Paint manufacturing
plant is aware that MACRS and DDB are both accelerated
depreciation methods; however, out of curiosity,
she wants to determine which one provides
the faster write-off in the first 3 years for a recently
purchased mixer that has a first cost of $300,000, a
5-year recovery period, and a $60,000 salvage
value. Determine which method yields the lower
book value and by how much after 3 years. The annual
MACRS depreciation rates are 20%, 32%, and
19.2% for years 1, 2, and 3, respectively.
16.31 Railroad cars used to transport coal from Wyoming
mines to Texas power plants cost $1.2 million and
have an estimated salvage value of $300,000. Develop
the depreciation and book value schedules for
the GDS MACRS method by using two methods on
a spreadsheet—the VDB function and the MACRS
rates. Are the book value series the same?
16.32 A 120-metric-ton telescoping crane that cost
$320,000 is owned by Upper State Power. Salvage
is estimated at $75,000. ( a ) Compare book values
for MACRS and standard SL depreciation over a
7-year recovery period. ( b ) Explain how the estimated
salvage is treated using MACRS.
16.33 Youngblood Shipbuilding Yard just purchased
$800,000 in capital equipment for ship repairing
functions on dry-docked ships. Estimated salvage
is $150,000 for any year after 5 years of use. Compare
the depreciation and book value for year 3 for
each of the following depreciation methods.
(a) GDS MACRS where a recovery period of
10 years is allowed
(b) Double declining balance with a recovery
period of 15 years
(c) ADS straight line as an alternative to MACRS,
with a recovery period of 15 years
16.34 Basketball.com has installed $100,000 worth of
depreciable software and equipment that represents
the latest in Internet teaming and basket
competition, intended to allow anyone to enjoy

Problems 441
the sport on the Web or in the alley. No salvage
value is estimated. The company can depreciate
using MACRS for a 5-year recovery period or opt
for the ADS alternate system over 10 years using
the straight line method. The SL rates require the
half-year convention; that is, only 50% of the regular
annual rate applies for years 1 and 11.
( a) Construct the book value curves for both
methods on one graph. Show hand or spreadsheet
computations as instructed.
( b) After 3 years of use, what percentage of the
$100,000 basis is removed for each method?
Compare the two percentages.
16.35 A company has purchased special-purpose equipment
for the manufacture of rubber products
(asset class 30.11 in IRS Publication 946) and expects
to use it predominately outside the United
States. In this case, the ADS alternative to
MACRS is required for tax depreciation purposes.
The manager wants to understand the difference
in yearly recovery rates for classical SL, MACRS,
and the ADS alternative to MACRS. Using a recovery
period of 3 years, except for the ADS alternative,
which requires a 4-year recovery with
half-year convention included, prepare a single
graph showing the annual recovery rates (in
percent) for the three methods.
Depletion
16.36 A coal mine purchased 3 years ago for $7 million
was estimated to contain 4,000,000 tons of coal.
During the past 3 years the amount of coal removed
was 21,000, 18,000, and 20,000 tons,
respectively. The gross income obtained in these 3
years was $257,000 for the first year, $320,000 for
the second year, and $340,000 for the third year.
Determine ( a ) the cost depletion allowance for
each year and ( b ) the percentage of the purchase
price depleted thus far.
16.37 NA Forest Resources purchased forest acreage
for $500,000 from which an estimated 200 million
board feet of lumber is recoverable. The
company will sell the lumber for $0.10 per board
foot. No lumber will be sold for the next 2 years
because an environmental impact statement
must be completed before harvesting can begin.
In years 3 to 10, however, the company expects
to remove 20 million board feet per year. The
inflation rate is 8%, and the company’s MARR
is 10%.
( a) Determine the depletion amount in year 2 by
the cost depletion method.
( b) Determine the depletion amount in year 5 by
the percentage depletion method.
16.38 A sand and gravel pit purchased for $900,000 is
expected to yield 50,000 tons of gravel and 80,000
tons of sand per year. The gravel will sell for
$6 per ton and the sand for $9 per ton.
(a) Determine the depletion charge according to
the percentage depletion method. The percentage
depletion rate for sand and gravel
is 5%.
(b) If taxable income is $100,000 for the year, is
this depletion charge allowed? If not, how
much is allowed?
16.39 Vesco Mineral Resources purchased mineral rights
to land in the foothills of the Santa Cristo mountains.
The cost of the purchase was $9 million.
Vesco originally thought that it would be able to
extract 200,000 tons of lignite from the land, but
further exploration revealed that 280,000 tons
could be economically removed. If the company
sold 20,000 tons in year 1 and 30,000 tons in year
2, what would the depletion charges be each year
according to the cost depletion method?
16.40 A company owns gold mining operations in the
United States, Australia, and South Africa. The
Colorado mine has the taxable income and sales
results summarized below. Determine the annual
percentage depletion amount for the gold mine.
Year
Taxable
Income, $
Sales,
Ounces
Average
Sales Price,
$ per Ounce
1 1,500,000 1800 1115
2 2,000,000 5500 1221
3 800,000 2300 1246
16.41 A highway construction company operates a
quarry. During the last 5 years, the amount extracted
each year was 60,000, 50,000, 58,000
60,000, and 65,000 tons. The mine is estimated to
contain a total of 2.5 million tons of usable stones
and gravel. The quarry land had an initial cost of
$3.2 million. The company had a per-ton gross
income of $30 for the first year, $25 for the second
year, $35 for the next 2 years, and $40 for the
last year.
(a) Determine the depletion charge each year,
using the larger of the values for the two depletion
methods. Assume all depletion amounts
are less than 50% of taxable income.
(b) Compute the percent of the initial cost that
has been written off in these 5 years, using
the depletion charges in part ( a ).
(c) If the quarry operation is reevaluated after
the first 3 years of operation and estimated to
contain a total of 1.5 million tons remaining,
rework parts ( a ) and ( b ).

442 Chapter 16 Depreciation Methods
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
16.42 All of the following types of real property are depreciable
except:
(a) Warehouses
(b) Land
(c) Office buildings
(d) Test facilities
16.43 According to information on the IRS website, a
taxpayer can take a depreciation deduction as long
as the property meets all of the following requirements
except:
(a) The taxpayer must own the property.
(b) The taxpayer must use the property in an
(c)
income-producing activity.
The taxpayer must use the property for personal
purposes.
(d) The property must have a determinable useful
life of more than 1 year.
16.44 A machine with a 5-year life has a first cost of
$20,000 and a $2000 salvage value. Its annual operating
cost is $8000 per year. According to the
classical straight line method, the depreciation
charge in year 2 is nearest to:
( a ) $2800 ( b ) $3600
( c ) $4500 ( d ) $5300
16.45 A machine with a 10-year life is MACRSdepreciated.
The machine has a first cost of $40,000
with a $5000 salvage value. Its annual operating
cost is $7000 per year, and d t for years 1, 2, and 3 is
10.00%, 18.00%, and 14.40%, respectively. The
depreciation charge in year 3 is nearest to:
( a ) $5800 ( b ) $7200
( c ) $8500 ( d ) $9300
16.46 A coal mine purchased for $5 million has enough
coal to operate for 10 years. The annual cost is expected
to be $200,000 per year. The coal is
expected to sell for $150 per ton, with annual production
expected to be 10,000 tons. Coal has a
depletion percentage rate of 10%. The depletion
charge for year 6 according to the percentage depletion
method would be closest to:
( a ) $75,000
( b ) $100,000
( c ) $125,000
( d ) $150,000
16.47 The depreciation charge for a 5-year, straight line
depreciated vehicle is $3000 in year 4. If the first
cost was $20,000, the salvage value used in the depreciation
calculation was closest to:
( a ) $0 ( b ) $2500
( c ) $5000 ( d ) $7500
16.48 The book value of an asset that was DDBdepreciated
over a 10-year period was $5832 at the
end of year 4. If the first cost of the asset was
$80,000, the salvage value that was used in the depreciation
calculation was closest to:
( a ) $0 ( b ) $2000
( c ) $5000 ( d ) $8000
16.49 For an asset that has B $100,000, S $40,000,
and a 10-year depreciable life, the book value at
the end of year 4 according to the MACRS method
would be closest to ( d t values for years 1, 2, 3, 4,
and 5 are 10.00%, 18.00%, 14.40%, 11.52%, and
9.22%, respectively):
( a ) $58,700 ( b ) $62,400
( c ) $53,900 ( d ) $46,100
16.50 An asset that was depreciated over a 5-year period
by the MACRS method had a BV of $33,025 at the
end of year 3. If the MACRS depreciation rates for
years 1, 2, and 3, were 0.20, 0.32, and 0.192, respectively,
the basis of the asset was closest to:
( a ) $158,000 ( b ) $172,000
( c ) $185,000 ( d ) $193,000
16.51 A lumber company purchased a tract of land for
$70,000 that contained an estimated 25,000 usable
trees. The value of the land was estimated at
$20,000. In the first year of operation, the lumber
company cut down 5000 trees. According to the
cost depletion method, the depletion deduction for
year 1 is closest to:
( a ) $2000 ( b ) $7000
( c ) $10,000 ( d ) $14,000
16.52 An asset with a first cost of $50,000 and an estimated
salvage value of $10,000 is depreciated by
the MACRS method. If its book value at the end of
year 3 is $21,850 and its market value is $25,850,
the total amount of depreciation charged against
the asset up to this time is closest to:
( a ) $18,850 ( b ) $21,850
( c ) $25,850 ( d ) $28,150
16.53 Under the General Depreciation System (GDS) of
asset classification, any asset that is not in a stated
class is automatically assigned a recovery period of:
( a ) 5 years (b) 7 years
( c ) 10 years ( d ) 15 years
16.54 All of the following statements about the Alternative
Depreciation System (ADS) are true except:
(a) The half-year convention applies.
(b) Salvage value is neglected.
(c) The recovery periods are shorter than in GDS.
( d) The straight line method is required.

Appendix Problems 443
APPENDIX PROBLEMS
Sum-of-Years-Digits Depreciation
16A.1 A European manufacturing company has new
equipment with a first cost of 12,000 euros, an
estimated salvage value of 2000 euros, and a
recovery period of 8 years. Use the SYD method
to tabulate annual depreciation and book value.
16A.2 Earthmoving equipment with a first cost of
$150,000 is expected to have a life of 10 years.
The salvage value is expected to be 10% of the
first cost. Calculate ( a ) by hand and ( b ) by
spreadsheet the depreciation charge and book
value for years 2 and 7 using the SYD method.
16A.3 If B $12,000, n 6 years, and S is estimated
at 15% of B , use the SYD method to determine
( a ) the book value after 3 years and ( b ) the rate
of depreciation and the depreciation amount in
year 4.
Unit-of-Production Depreciation
16A.4 A robot used in simulated car crashes cost
$70,000, has no salvage value, and has an
expected capacity of tests not to exceed 10,000
according to the manufacturer. Volvo Motors decided
to use the unit-of-production depreciation
method because the number of test crashes per
year in which the robot would be involved was
not estimable. Determine the annual depreciation
and book value for the first 3 years if the number
of tests were 3810, 2720, and 5390 per year.
16A.5 A new hybrid car was purchased by Pedernales
Electric Cooperative as a courier vehicle to
transport items between its 12 city offices. The
car cost $30,000 and was retained for 5 years.
Alternatively, it could have been retained for
100,000 miles. Salvage value is nil. Five-year
DDB depreciation was applied. The car pool
manager stated that he prefers UOP depreciation
on vehicles because it writes off the first
cost faster. Use the actual annual miles driven,
listed below, to plot the book values for both
methods. Determine which method would have
removed the $30,000 faster. Show hand or
spreadsheet solution, as instructed.
Year 1 2 3 4 5
Miles Driven, 1000 15 22 16 18 24
Switching Methods
16A.6 An asset has a first cost of $45,000, a recovery
period of 5 years, and a $3000 salvage value.
Use the switching procedure from DDB to SL
depreciation, and calculate the present worth of
depreciation at i 18% per year.
16A.7 If B $45,000, S $3000, and n 5-year
recovery period, use a spreadsheet and i
18% per year to maximize the present worth of
depreciation, using the following methods:
DDB-to-SL switching (this was determined in
Problem 16A.6) and MACRS. Given that
MACRS is the required depreciation system in
the United States, comment on the results.
16A.8 Hempstead Industries has a new milling machine
with B $110,000, n 10 years, and S $10,000.
Determine the depreciation schedule and present
worth of depreciation at i 12% per year, using
the 175% DB method for the first 5 years and
switching to the classical SL method for the last 5
years. Use a spreadsheet to solve this problem.
16A.9 Reliant Electric Company has erected a large
portable building with a first cost of $155,000
and an anticipated salvage of $50,000 after
25 years. ( a ) Should the switch from DDB to
SL depreciation be made? ( b ) For what values
of the uniform depreciation rate in the DB
method would it be advantageous to switch
from DB to SL depreciation at some point in
the life of the building?
MACRS Rates
16A.10 Verify the 5-year recovery period rates for
MACRS given in Table 16–2. Start with the
DDB method in year 1, and switch to SL depreciation
when it offers a larger recovery rate.
16A.11 A video recording system was purchased
3 years ago at a cost of $30,000. A 5-year recovery
period and MACRS depreciation have been used
to write off the basis. The system is to be prematurely
replaced with a trade-in value of $5000.
Determine the MACRS depreciation, using the
switching rules to find the difference between the
book value and the trade-in value after 3 years.
16A.12 Use the computations in Equations [16A.11]
through [16A.13] to determine the MACRS annual
depreciation for the following asset data:
B $50,000 and a recovery period of 7 years.
16A.13 The 3-year MACRS recovery rates are 33.33%,
44.45%, 14.81%, and 7.41%, respectively.
( a ) What are the corresponding rates for the alternative
MACRS straight line ADS method
with the half-year convention imposed?
( b ) Compare the PW D values for these two methods
if B $80,000 and i 15% per year.

CHAPTER 17
After-Tax
Economic
Analysis
LEARNING OUTCOMES
Purpose: Perform an after-tax economic evaluation considering the impact of pertinent tax regulations, income taxes,
and depreciation.
SECTION TOPIC LEARNING OUTCOME
17.1 Terminology and rates • Know the fundamental terms and relations of after-tax
analysis; use a marginal tax rate table.
17.2 CFBT and CFAT • Determine cash flow series before taxes and after taxes.
17.3 Taxes and depreciation • Demonstrate the tax advantage of accelerated
depreciation and shortened recovery periods.
17.4 Depreciation recapture • Calculate the tax impact of DR; explain the use of
capital gains and capital losses.
17.5 After-tax analysis • Evaluate one project or multiple alternatives using
after-tax PW, AW, and ROR analysis.
17.6 After-tax replacement • Evaluate a defender and challenger in an after-tax
replacement study.
17.7 EVA analysis • Evaluate an alternative using after-tax economic valueadded
analysis; compare to CFAT analysis.
17.8 Taxes outside the United States • Understand the fundamental practices for depreciation
and tax rates in international settings.
17.9 VAT • Demonstrate the use and computation of a valueadded
tax on manufactured products.

T
his chapter provides an overview of tax terminology, income tax rates, and tax
equations pertinent to an after-tax economic analysis. The change from estimating
cash flow before taxes (CFBT) to cash flow after taxes (CFAT) involves a consideration
of significant tax effects that may alter the final decision, and estimates the magnitude
of the effect on cash flow over the life of the alternative that taxes may have.
Mutually exclusive alternative comparisons using after-tax PW, AW, and ROR methods
are explained with major tax implications considered. Replacement studies are discussed
with tax effects that occur at the time that a defender is replaced. Also, the after-tax economic
value added by an alternative is discussed in the context of annual worth analysis. All
these methods use the procedures learned in earlier chapters, except now with tax effects
considered.
An after-tax evaluation using any method requires more computations than in previous
chapters. Templates for tabulation of cash flow after taxes by hand and by spreadsheet
are developed. Additional information on U.S. federal taxes—tax law and annually updated
tax rates—is available through Internal Revenue Service publications and, more readily,
on the IRS website www.irs.gov. Publications 542, Corporations , and 544, Sales and Other
Dispositions of Assets , are especially applicable to this chapter. Some differences in tax considerations
outside the United States are summarized.
17.1 Income Tax Terminology and Basic Relations
The perspective taken in engineering economy when performing an after-tax evaluation is that of
the project and how relevant tax rules and allowances influence the economic decision. The perspective
of a financial study is that of the corporation and how the tax structure and laws affect
profitability. We take the engineering economy viewpoint in the sections that follow.
There are many types of taxes levied upon corporations and individuals in all countries, including
the United States. Some are sales tax, valued-added tax, import tax, income tax, highway
tax, gasoline tax, and property (real estate) tax. Federal governments rely on income taxes for a
significant portion of their annual revenue. States, provinces, and municipal-level governments
rely on sales, value-added, and property taxes to maintain services, schools, etc. for the citizenry.
For a starting point on corporate income taxes and how they are used when performing an aftertax
economic evaluation of a project or multiple alternatives, this section covers basic definitions,
terms, and relations.
Income tax is the amount of the payment (taxes) on income or profit that must be delivered to a
federal (or lower-level) government unit. Taxes are real cash flows ; however, for corporations
tax computation requires some noncash elements, such as depreciation. Corporate income taxes
are usually submitted quarterly, and the last payment of the year is submitted with the annual tax
return.
The Internal Revenue Service (IRS), a part of the U.S. Department of the Treasury, collects the
taxes and enforces tax laws. The website www.irs.gov provides information on tax laws, rates,
publications, etc. that are referenced in this chapter.
Though the formulas are much more complex when applied to a specific situation, two fundamental
relations form the basis for income tax computations. The first involves only actual cash
flows:
Net operating income revenue operating expenses
The second involves actual cash flows and noncash deductibles, such as depreciation.
Taxable income revenue operating expenses depreciation
These terms and relations for corporations are now described. Since each term is calculated for
1 year, there can be a subscript t ( t 1, 2, . . .) added; t is omitted here for simplicity.
Operating revenue R , also commonly called gross income GI , is the total income realized
from all revenue-producing sources. These incomes are listed in the income statement. (See
Appendix B on accounting reports.) Other, nonoperating revenues such as sale of assets,
license fee income, and royalties are considered separately for tax purposes.

446 Chapter 17 After-Tax Economic Analysis
Operating expenses OE include all costs incurred in the transaction of business. These expenses
are tax-deductible for corporations. For after-tax economic evaluations, the AOC (annual
operating costs) and M&O (maintenance and operating) costs are applicable here. Depreciation
is not included here since it is not an operating expense .
Net operating income NOI, often called EBIT (earnings before interest and income taxes),
is the difference between gross income and operating expenses.
NOI EBIT GI OE [17.1]
Taxable income TI is the amount of income upon which taxes are based. A corporation is
allowed to remove depreciation , depletion and amortization, and some other deductibles
from net operating income in determining the taxable income for a year. For our evaluations,
we define taxable income as
TI gross income operating expenses depreciation
GI OE D [17.2]
Though there may be subtleties and varying interpretations over time, in essence, the differences
between NOI and TI are tax-law-allowed deductibles, such as depreciation. (In keeping
with the project view of engineering economics, we will primarily use the TI relation when
conducting an after-tax evaluation.)
Tax rate T is a percentage, or decimal equivalent, of TI that is owed in taxes. The tax rates
in many countries (including the United States) are graduated (or progressive) by level of
TI; that is, higher rates apply as the TI increases. The marginal tax rate is the percentage
paid on the last dollar of income . The average tax rate paid is calculated separately from the
highest marginal rate used, as shown later. The general tax computation relation is
Income taxes applicable tax rate taxable income
( T )(TI) [17.3]
Net operating profit after taxes NOPAT is the amount remaining each year after taxes are
subtracted from taxable income.
NOPAT TI taxes TI ( T )(TI)
TI(1 T) [17.4]
Basically, NOPAT represents the money remaining in the corporation as a result of the capital
invested during the year. It is also called net profi t after taxes (NPAT).
The graduated tax rate schedule for corporations is presented in Table 17–1 as taken from IRS
Publication 542, Corporations . These are rates for the entire corporation, not for an individual
project, though they are often applied in the after-tax analysis of a single project. The rates can
change annually based upon government legislation; however, the corporate tax rate schedule has
remained the same for some years. To illustrate the use of the graduated tax rate, assume a company
is expected to generate a taxable income of $500,000 in 1 year. From Table 17–1 , the marginal
tax rate for the last dollar of TI is 34%, but the graduated rates become progressively larger
as TI increases. In this case, for TI $500,000,
Taxes 113,900 0.34(500,000 335,000)
113,900 56,100
$170,000
Alternatively, the rates for each TI level can be used to calculate taxes the longer way.
Taxes 0.15(50,000) 0.25(75,000 50,000) 0.34(100,000 75,000)
0.39(335,000 100,000) 0.34(500,000 335,000)
7500 6250 8500 91,650 56,100
$170,000

17.1 Income Tax Terminology and Basic Relations 447
TABLE 17–1 U.S. Corporate Income Tax Rate Schedule (2010)
Over
But Not
over
If Taxable Income ($) Is:
Tax Is
Of the
Amount over
0 50,000 15% 0
50,000 75,000 7,500 25% 50,000
75,000 100,000 13,750 34% 75,000
100,000 335,000 22,250 39% 100,000
335,000 10,000,000 113,900 34% 335,000
10,000,000 15,000,000 3,400,000 35% 10,000,000
15,000,000 18,333,333 5,150,000 38% 15,000,000
18,333,333 — 35% 0
Smaller businesses (with TI $335,000) receive a slight tax advantage compared to large corporations.
Once the TI exceeds $335,000, an effective federal tax rate of 34% applies, and when
TI $18.33 million, there is a flat tax rate of 35%.
As we move forward with after-tax analysis, it is important to keep the following in mind.
The corporate tax rates apply to a corporation as a whole, not to a specific project, unless the
project is the company. Tax rates are usually graduated by level of taxable income. Therefore, the
last dollar of TI is taxed at a marginal rate. The average federal tax rate actually paid is lower than
the highest marginal rate paid because the rates, in general, graduate to higher percentages as
TI increases.
Because the marginal tax rates change with TI, it is not possible to quote directly the percent
of TI paid in income taxes. Alternatively, a single-value number, the average tax rate, is
calculated as
total taxes paid
Average tax rate ———————
taxable income ——— taxes
[17.5]
TI
Referring to Table 17–1 , for a small business with TI $100,000, the federal income tax
burden averages $22,250100,000 22.25%. If TI $15 million, the average tax rate is
$5.15 million15 million 34.33%.
As mentioned earlier, there are federal, state, and local taxes imposed. For the sake of simplicity,
the tax rate used in an economy study is often a single-figure effective tax rate T e , which
accounts for all taxes. Effective tax rates are in the range of 35% to 50%. One reason to use the
effective tax rate is that state taxes are deductible for federal tax computation. The effective tax
rate and taxes are calculated as
EXAMPLE 17.1
T e state rate (1 state rate)(federal rate) [17.6]
Taxes ( T e )(TI) [17.7]
REI (Recreational Equipment Incorporated) sells outdoor equipment and sporting goods through
retail outlets, the Internet, and catalogs. Assume that for 1 year REI has the following financial
results in the state of Kentucky, which has a flat tax rate of 6% on corporate taxable income.
Total revenue
$19.9 million
Operating expenses
$8.6 million
Depreciation and other allowed deductions $1.8 million
(a) Determine the state taxes and federal taxes due using Table 17–1 rates.
(b) Find the average federal tax rate paid for the year.
(c) Determine a single-value tax rate useful in economic evaluations using the average federal
tax rate determined in part (b).
(d) Estimate federal and state taxes using the single-value rate, and compare their total with the
total in part (a).

448 Chapter 17 After-Tax Economic Analysis
Solution
(a) Calculate TI by Equation [17.2] and use Table 17–1 rates for federal taxes due.
Kentucky state TI GI OE D 19.9 million 8.6 million 1.8 million
$9.5 million
Kentucky state taxes 0.06(TI) 0.06(9,500,000) $570,000
Federal TI GI OE D state taxes 9,500,000 570,000
$8,930,000
Federal taxes 113,900 0.34(8,930,000 335,000) $3,036,200
Total federal and state taxes 3,036,200 570,000 $3,606,200 [17.8]
(b) From Equation [17.5], the average tax rate paid is approximately 32% of TI.
Average federal tax rate 3,036,2009,500,000 0.3196
(c) By Equation [17.6], T e is slightly over 36% per year for combined state and federal taxes.
T e 0.06 (1 0.06)(0.3196) 0.3604 (36.04%)
(d) Use the effective tax rate and TI $9.5 million from part (a) in Equation [17.7] to
approximate total taxes.
Taxes 0.3604(9,500,000) $3,423,800
Compared to Equation [17.8], this approximation is $182,400 low, a 5.06% underestimate.
It is interesting to understand how corporate tax and individual tax computations differ. Gross
income for an individual taxpayer is comparable if revenue is replaced by salaries and wages.
However, for an individual’s taxable income, most of the expenses for living and working are not
tax deductible to the same degree as operating expenses are for corporations. For individual taxpayers,
GI salaries wages interest and dividends other income
TI GI personal exemption standard or itemized deductions
Taxes ( T )(TI)
For TI, operating expenses are replaced by individual exemptions and specific deductions. Exemptions
are yourself, your spouse, your children, and your other dependents. Each exemption
reduces TI by $3500 to $4000 per year, depending upon current exemption allowances.
In the United States, the tax rates for individuals, like those for corporations, are graduated by
level of TI. In 2010, the marginal rates ranged from 10% to 35%; however, the top marginal rates
are increasing for individuals with larger TI amounts. These rates and TI levels are the subject of
ongoing political debates at the U.S. national level depending upon the balance of power in the
congressional bodies and the economic condition of the country. Once the marginal rates are set,
the TI levels are adjusted each year to account for inflation and other factors. This process is
called indexing . Clearly, tax rates for individuals change much more frequently than the rates for
corporations change. Current information is available on the IRS website www.irs.gov through
Publication 17, Your Federal Income Tax. The current rate schedule is published in the back of
this document for four filing status categories:
Unmarried individuals (single)
Married filing jointly
Married filing separately
Head of household
17.2 Calculation of Cash Flow after Taxes
Early in the text, the term net cash fl ow ( NCF ) was identified as the best estimate of actual cash
flow each year. The NCF is calculated as cash inflows minus cash outflows. Since then, the annual
NCF amounts have been used many times to perform alternative evaluations via the PW,
AW, ROR, and B/C methods. Now that the impact on cash flow of depreciation and related taxes

17.2 Calculation of Cash Flow after Taxes 449
will be considered, it is time to expand our terminology. NCF is replaced by the term cash flow
before taxes (CFBT) , and we introduce the new term cash flow after taxes (CFAT) .
CFBT and CFAT are actual cash flows ; that is, they represent the estimated actual flow of
money into and out of the corporation that will result from the alternative. The remainder of
this section explains how to transition from before-tax to after-tax cash flows for solutions by
hand and by spreadsheet, using tax regulations described in the next few sections. Once the
CFAT estimates are developed, the economic evaluation is performed using the same methods
and selection guidelines applied previously. However, the analysis is performed on the CFAT
estimates.
We learned that net operating income (NOI) does not include the purchase or sale of capital
assets. However, the annual CFBT estimate must include the initial capital investment and salvage
value for the years in which they occur. Incorporating the definitions of gross income and
operating expenses from NOI, CFBT for any year is defined as
CFBT gross income operating expenses initial investment salvage value
GI OE P S [17.9]
As in previous chapters, P is the initial investment (year 0) and S is the estimated salvage value
in year n. Therefore, only in year 0 will the CFBT include P , and only in year n will an S value
be present. Once all taxes are estimated, the annual after-tax cash flow is simply
CFAT CFBT taxes [17.10]
where taxes are estimated using the relation ( T )(TI) or ( T e )(TI).
We know from Equation [17.2] that depreciation D is considered when calculating TI. It is
very important to understand the different roles of depreciation for income tax computations and
in CFAT estimation.
Depreciation is not an operating expense and is a noncash flow. Depreciation is tax deductible for
determining the amount of income taxes only, but it does not represent a direct, after-tax cash flow.
Therefore, the after-tax engineering economy study must be based on actual cash flow estimates,
that is, annual CFAT estimates that do not include depreciation as an expense (negative cash flow).
Accordingly, if the CFAT expression is determined using the TI relation, depreciation must not
be included outside of the TI component. Equations [17.9] and [17.10] are now combined.
CFAT GI OE P S (GI OE D )( T e ) [17.11]
Suggested table column headings for CFBT and CFAT calculations by hand or by spreadsheet are
shown in Table 17–2 . The equations are shown in column numbers, with the effective tax rate T e
used for income tax estimation. Expenses OE and initial investment P carry negative signs in all
tables and spreadsheets.
A negative TI value may occur in some years due to a depreciation amount that is larger than
(GI − OE). It is possible to account for this in a detailed after-tax analysis using carry-forward
and carry-back rules for operating losses. It is the exception that the engineering economy study
will consider this level of detail. Rather, the associated negative income tax is considered as a tax
savings for the year . The assumption is that the negative tax will offset taxes for the same year
in other income-producing areas of the corporation.
TABLE 17–2
Suggested Column Headings for Calculation of CFAT
Year
Gross
Income
GI
Operating
Expenses
OE
Investment
and
Salvage
P and S
CFBT
(1) (2) (3) (4)
(1) (2) (3)
Depreciation
D
(5) (6)
(1) (2) (5)
Taxable
Income
TI Taxes CFAT
(7)
T e (6)
(8)
(4) (7)

450 Chapter 17 After-Tax Economic Analysis
EXAMPLE 17.2
Wilson Security has received a contract to provide additional security for corporate and government
personnel along the Arizona-Mexico border. Wilson plans to purchase listening and detection
equipment for use in the 6-year contract. The equipment is expected to cost $550,000 and have a
resale value of $150,000 after 6 years. Based on the incentive clause in the contract, Wilson estimates
that the equipment will increase contract revenue by $200,000 per year and require an additional
M&O expense of $90,000 per year. MACRS depreciation allows recovery in 5 years, and
the effective corporate tax rate is 35% per year. Tabulate and plot the CFBT and CFAT series.
Solution
The spreadsheet in Figure 17–1 presents before-tax and after-tax cash flows using the format
of Table 17–2. The functions for year 6 are detailed in row 11. Discussion and sample calculations
follow.
CFBT: The operating expenses OE and initial investment P are shown as negative cash flows.
The $150,000 salvage (resale) is a positive cash flow in year 6. CFBT is calculated by Equation
[17.9]. In year 6, for example, when the equipment is sold, the function in row 11 indicates that
CFBT 6 200,000 90,000 150,000 $260,000
CFAT: Column F for MACRS depreciation, which is determined using the VDB function
over the 6-year period, writes off the entire $550,000 investment. Taxable income, taxes,
and CFAT are calculated, using year 4 as an example.
TI 4 GI OE D 200,000 90,000 63,360 $46,640
Taxes 4 (0.35) (TI) (0.35) (46,640) $16,324
CFAT 4 GI OE taxes 200,000 − 90,000 16,324 $93,676
In year 2, MACRS depreciation is large enough to cause TI to be negative ($−66,000). As
mentioned above, the negative tax ($−23,100) is considered a tax savings in year 2, thus
increasing CFAT.
Comment
MACRS depreciates to a salvage value of S 0. Later we will learn about a tax implication
due to “recapturing of depreciation” when an asset is sold for an amount larger than zero and
MACRS was applied to fully depreciate the asset to zero.
Figure 17–1
Computation of CFBT and CFAT using MACRS depreciation and T e 35%, Example 17.2.
17.3 Effect on Taxes of Different Depreciation
Methods and Recovery Periods
It is important to understand why accelerated depreciation rates give the corporation a tax advantage
relative to the straight line method with the same recovery period. Larger rates in earlier
years of the recovery period require less taxes due to the larger reductions in taxable income. The
criterion of minimizing the present worth of taxes is used to demonstrate the tax effect. For the

17.3 Effect on Taxes of Different Depreciation Methods and Recovery Periods 451
recovery period n , choose the depreciation rates that result in the minimum present worth value
for taxes.
tn
PW tax (taxes in year t)(PF, i, t) [17.12]
t1
This is equivalent to maximizing the present worth of total depreciation PW D .
Compare any two depreciation methods. Assume the following: (1) There is a constant singlevalue
tax rate, (2) CFBT exceeds the annual depreciation amount, (3) both methods reduce book
value to the same salvage value, and (4) the same recovery period is used. On the basis of these
assumptions, the following statements are correct:
The total taxes paid are equal for all depreciation methods.
The present worth of taxes is less for accelerated depreciation methods.
As we learned in Chapter 16, MACRS is the prescribed tax depreciation method in the United
States, and the only alternative is MACRS straight line depreciation with an extended recovery
period. The accelerated write-off of MACRS always provides a smaller PW tax compared to less
accelerated methods. If the DDB method were still allowed directly, rather than embedded in
MACRS, DDB would not fare as well as MACRS. This is so because DDB does not reduce the
book value to zero. This is illustrated in Example 17.3 .
EXAMPLE 17.3
An after-tax analysis for a new $50,000 machine proposed for a fiber optics manufacturing line
is in process. The CFBT for the machine is estimated at $20,000. If a recovery period of 5 years
applies, use the present worth of taxes criterion, an effective tax rate of 35%, and a return of
8% per year to compare the following: classical straight line, classical DDB, and required
MACRS depreciation. Use a 6-year period for the comparison to accommodate the half-year
convention imposed by MACRS.
Solution
Table 17–3 presents a summary of annual depreciation, taxable income, and taxes for each
method. For classical straight line depreciation with n 5, D t $10,000 for 5 years and D 6
0 (column 3). The CFBT of $20,000 is fully taxed at 35% in year 6.
The classical DDB percentage of d 2n 0.40 is applied for 5 years. The implied salvage
value is $50,000 − 46,112 $3888, so not all $50,000 is tax-deductible. The taxes using classical
DDB will be $3888(0.35) $1361 larger than for the classical SL method.
TABLE 17–3
Comparison of Taxes and Present Worth of Taxes for Different Depreciation Methods
(1)
Year t
(2)
CFBT, $
(3)
D t , $
Classical Straight Line
(4)
TI, $
(5) 0.35(4)
Taxes, $
Classical Double Declining
Balance
(6)
D t , $
(7)
TI, $
(8) 0.35(7)
Taxes, $
(9)
D t , $
(10)
TI, $
MACRS
(11) 0.35(10)
Taxes, $
1 20,000 10,000 10,000 3,500 20,000 0 0 10,000 10,000 3,500
2 20,000 10,000 10,000 3,500 12,000 8,000 2,800 16,000 4,000 1,400
3 20,000 10,000 10,000 3,500 7,200 12,800 4,480 9,600 10,400 3,640
4 20,000 10,000 10,000 3,500 4,320 15,680 5,488 5,760 14,240 4,984
5 20,000 10,000 10,000 3,500 2,592 17,408 6,093 5,760 14,240 4,984
6 20,000 0 20,000 7,000 0 20,000 7,000 2,880 17,120 5,992
Totals 50,000 24,500 46,112 25,861* 50,000 24,500
PW tax 18,386 18,549 18,162
*Larger than other values since there is an implied salvage value of $3888 not recovered.

452 Chapter 17 After-Tax Economic Analysis
25,000
20,000
Accumulated taxes, $
15,000
10,000
MACRS
method
SL
method
DDB method
5,000
0 1
2 3
Year, t
4 5 6
Figure 17–2
Cumulative taxes incurred by different depreciation rates for a
6-year comparison period, Example 17.3.
MACRS writes off $50,000 in 6 years using the rates of Table 16–2. Total taxes are $24,500,
the same as for classical SL depreciation.
The annual taxes (columns 5, 8, and 11) are accumulated year by year in Figure 17–2. Note
the pattern of the curves, especially the lower total taxes relative to the SL model after year 1
for MACRS and in years 1 through 4 for DDB. These higher tax values for SL cause PW tax for
SL depreciation to be larger. The PW tax values at the bottom of Table 17–3 are calculated using
Equation [17.12]. The MACRS PW tax value is the smallest at $18,162.
To compare taxes for different recovery periods , change only assumption 4 at the beginning
of this section to read: The same depreciation method is applied. It can be shown that a shorter
recovery period will offer a tax advantage over a longer period using the criterion to minimize
PW tax . Comparison will indicate that
The total taxes paid are equal for all n values.
The present worth of taxes is less for smaller n values.
This is why corporations want to use the shortest MACRS recovery period allowed for income
tax purposes. Example 17.4 demonstrates these conclusions for classical straight line depreciation,
but the conclusions are correct for MACRS or any other tax depreciation method.
EXAMPLE 17.4
Grupo Grande Maquinaría, a diversified manufacturing corporation based in Mexico, maintains
parallel records for depreciable assets in its European operations in Germany. This is
common for multinational corporations. One set is for corporate use that reflects the estimated
useful life of assets. The second set is for foreign government purposes, such as
depreciation and taxes.
The company just purchased an asset for $90,000 with an estimated useful life of 9 years;
however, a shorter recovery period of 5 years is allowed by German tax law. Demonstrate the
tax advantage for the smaller n if net operating income (NOI) is $30,000 per year, an effective
tax rate of 35% applies, invested money is returning 5% per year after taxes, and classical SL
depreciation is allowed. Neglect the effect of any salvage value.
Solution
Determine the annual TI and taxes by Equations [17.2] through [17.3] and the present worth of
taxes using Equation [17.12] for both n values.

17.4 Depreciation Recapture and Capital Gains (Losses) 453
Useful life n 9 years:
D ———
90,000 10,000
9
TI 30,000 10,000 $20,000 per year
Taxes (0.35)(20,000) $7000 per year
PW tax 7000(PA,5%,9) $49,755
Total taxes (7000)(9) $63,000
Recovery period n 5 years:
Use the same comparison period of 9 years, but depreciation occurs only during the first 5 years.
———
D t
{ 90,000 $18,000 t 1 to 5
5
0 t 6 to 9
18,000) $4200 t 1 to 5
Taxes {(0.35)(30,000) $10,500 t 6 to 9
PW tax 4200(PA,5%,5) 10,500(PA,5%,4)(PF,5%,5)
$47,356
Total taxes 4200(5) 10,500(4) $63,000
A total of $63,000 in taxes is paid in both cases. However, the more rapid write-off for
n 5 results in a present worth tax savings of nearly $2400 (49,755 47,356).
17.4 Depreciation Recapture and Capital
Gains (Losses)
All the tax implications discussed here are the result of disposing of a depreciable asset before, at,
or after its recovery period. In an after-tax economic analysis of large investment assets, these tax
effects should be considered. The key is the size of the selling price (or salvage or market value)
relative to the current book value at disposal time and relative to the first cost, which is called the
unadjusted basis B in depreciation terminology. There are three relevant terms.
Depreciation recapture DR , also called ordinary gain , occurs when a depreciable asset is
sold for more than the current book value BV t . As shown in Figure 17–3 ,
Depreciation recapture selling price book value
DR SP 2 BV t [17.13]
If selling price
(SP) at time of
sale is:
The capital gain,
depreciation recapture,
or capital loss is:
For an after-tax
study, the tax
effect is:
First cost P
or basis B
SP 1
CG
plus
CG: Taxed at T e
(after CL offset)
DR
DR: Taxed at T e
Book
value BV t
SP 2
SP 3
CL
DR
CL: Can only offset
CG
Zero, $0
Figure 17–3
Summary of calculations and tax treatment for depreciation recapture and capital gains (losses).

454 Chapter 17 After-Tax Economic Analysis
Depreciation recapture is often present in the after-tax study . In the United States, an amount
equal to the estimated salvage value can always be anticipated as DR when the asset is disposed
of after the MACRS recovery period. This is correct simply because MACRS depreciates
every asset to zero in n 1 years. The amount of DR is treated as ordinary taxable income
in the year of asset disposal.
Capital gain CG is an amount incurred when the selling price exceeds its (unadjusted) basis
B . See Figure 17–3 .
Capital gain selling price basis
CG SP 1 B [17.14]
Since future capital gains are difficult to predict, they are usually not detailed in an after-tax economy
study. An exception is for assets that historically increase in value, such as buildings and land.
When the selling price exceeds B , the TI due to the sale is the capital gain plus the depreciation
recapture, as shown in Figure 17–3 . The DR is now the total amount of depreciation
taken thus far, that is, B BV .
Capital loss CL occurs when a depreciable asset is disposed of for less than its current book
value. In Figure 17–3 ,
Capital loss book value selling price
CL BV t SP 3 [17.15]
An economic analysis does not commonly account for capital loss, simply because it is not
estimable for a specific alternative. However, an after-tax replacement study should account
for any capital loss if the defender must be traded at a “sacrifice” price. For the purposes of the
economic study, this provides a tax savings in the year of replacement. Use the effective tax
rate to estimate the tax savings. These savings are assumed to be offset elsewhere in the corporation
by other income-producing assets that generate taxes.
There are several additional points worth mentioning about capital gains and capital losses for
a corporation, apart from their presence in an economic evaluation.
• U.S. tax law defines capital gains as long-term (items retained for more than 1 year) or short-term.
• Capital gains actually take place for property that is not depreciated or amortized. The term
capital gain correctly applies at sale time to property such as investments (stocks and bonds),
art, jewelry, land, and the like. When a depreciable asset’s selling price is higher than the
original cost (its basis), it is correctly termed an ordinary gain . Corporate tax treatment at this
time is the same for both: taxed as ordinary income. All said, it is common to classify an
expected ordinary gain on a depreciable asset as a capital gain, without altering the economic
decision.
• Capital gains are taxed as ordinary taxable income at the corporation’s regular tax rates.
• Capital losses do not directly reduce annual income taxes because they can only be netted
against capital gains to the maximum extent of the capital gains for the year. The terms used
then are net capital gains (losses).
• When capital losses exceed capital gains, the corporation can take advantage of carry-back
and carry-forward tax laws for the excess, something of value to a finance or tax officer, not
an engineer doing an economic analysis.
• When an asset is disposed of, the tax treatment is referred to as a Section 1231 transaction ,
which is the IRS rule section of the same number.
• IRS Publication 544, Sales and Other Dispositions of Assets , may be helpful if gains and
losses are present in a study.
• All these rules apply to corporations. Individual tax rules and rates are different concerning
asset disposition.
If the three additional income and tax elements covered here are incorporated into Equation
[17.2], taxable income is defined as
TI gross income operating expenses depreciation
depreciation recapture net capital gain net capital loss
GI OE D DR CG CL [17.16]

17.4 Depreciation Recapture and Capital Gains (Losses) 455
In keeping with our perspective of an engineering economy study rather than that of a financial
study, deprecation recapture (i.e., ordinary gain) will be the primary element considered in aftertax
evaluations. Only when a capital gain or loss must be included due to the nature of the problem
will the calculations involve it.
EXAMPLE 17.5
Biotech, a medical imaging and modeling company, must purchase a bone cell analysis system for
use by a team of bioengineers and mechanical engineers studying bone density in athletes. This
particular part of a 3-year contract with the NBA will provide additional gross income of $100,000
per year. The effective tax rate is 35%. Estimates for two alternatives are summarized below.
Analyzer 1 Analyzer 2
Basis B, $ 150,000 225,000
Operating expenses, $ per year 30,000 10,000
MACRS recovery, years 5 5
Answer the following questions, solving by hand and spreadsheet.
(a) The Biotech president, who is very tax conscious, wishes to use a criterion of minimizing
total taxes incurred over the 3 years of the contract. Which analyzer should be purchased?
(b) Assume that 3 years have now passed, and the company is about to sell the analyzer. Using
the same total tax criterion, did either analyzer have an advantage? Assume the selling
price is $130,000 for analyzer 1, or $225,000 for analyzer 2.
Solution by Hand
(a) Table 17–4 details the tax computations. First, the yearly MACRS depreciation is determined.
Equation [17.2], TI GI − OE − D, is used to calculate TI, after which the 35% tax
rate is applied each year. Taxes for the 3-year period are summed, with no consideration of
the time value of money.
Analyzer 1 tax total: $36,120 Analyzer 2 tax total: $38,430
The two analyzers are very close, but analyzer 1 wins with $2310 less in total taxes.
(b) When the analyzer is sold after 3 years of service, there is a depreciation recapture (DR)
that is taxed at the 35% rate. This tax is in addition to the third-year tax. For each analyzer,
account for the DR by Equation [17.13], SP BV 3 ; then determine the TI, using
Equa tion [17.16], TI GI − OE − D DR. Again, find the total taxes over 3 years, and
select the analyzer with the smaller total.
TABLE 17–4 Comparison of Total Taxes for Two Alternatives, Example 17.5a
Year
Gross
Income
GI, $
Operating
Expenses
OE, $
Basis
B, $
MACRS
Depreciation
D, $
Analyzer 1
Book
Value
BV, $
Taxable
Income
TI, $
Taxes
at
0.35TI, $
0 150,000 150,000
1 100,000 30,000 30,000 120,000 40,000 14,000
2 100,000 30,000 48,000 72,000 22,000 7,700
3 100,000 30,000 28,800 43,200 41,200 14,420
36,120
Analyzer 2
0 225,000 225,000
1 100,000 10,000 45,000 180,000 45,000 15,750
2 100,000 10,000 72,000 108,000 18,000 6,300
3 100,000 10,000 43,200 64,800 46,800 16,380
38,430

456 Chapter 17 After-Tax Economic Analysis
Analyzer 1: DR 130,000 43,200 $86,800
Year 3 TI 100,000 30,000 28,800 86,800 $128,000
Year 3 taxes (0.35)(128,000) $44,800
Total taxes 14,000 7700 44,800 $66,500
Analyzer 2: DR 225,000 64,800 $160,200
Year 3 TI 100,000 10,000 43,200 160,200 $207,000
Year 3 taxes (0.35)(207,000) $72,450
Total taxes 15,750 6300 72,450 $94,500
Now, analyzer 1 has a considerable advantage in total taxes ($94,500 versus $66,500).
Solution by Spreadsheet
(a) Rows 5 through 9 of Figure 17–4 perform the same computations as the hand solution for
analyzer 1 with total taxes of $36,120. Similar analysis in rows 14 to 18 results in total
taxes of $38,430 for analyzer 2, indicating that the company should select analyzer 1, based
on taxes only.
(b) Revised year 3 entries for analyzer 1 in row 10 show the sales price of $130,000, the updated
TI of $128,000, and a 3-year tax total of $66,500. The new TI in year 3 has the depreciation
recapture incorporated as DR selling price book value SP BV 3 , which
is shown in the cell tag as the last term (D10 F10). With a similar updating for analyzer
2 (row 19), total taxes of $94,500 now show a significantly larger tax advantage for analyzer
1 over 3 years.
Comment
Note that no time value of money is considered in these analyses, as we have used in previous
alternative evaluations. In Section 17.5 below we will rely upon PW, AW, and ROR analyses
at an established MARR to make an after-tax decision based upon CFAT values.
B6C6E6
B10C10E10(D10F10)
Last term calculates the
depreciation recapture
Total tax advantage
of analyzer 1 over 2
without sale and
with sale in year 3
Figure 17–4
Impact of depreciation recapture on total taxes, Example 17.5.
17.5 After-Tax Evaluation
The required after-tax MARR is established using the market interest rate, the corporation’s effective
tax rate, and its weighted average cost of capital. The CFAT estimates are used to compute
the PW or AW at the after-tax MARR. When positive and negative CFAT values are present, a
PW or AW 0 indicates the MARR is not met. For a single project or mutually exclusive alternatives,
apply the same logic as in Chapters 5 and 6. The guidelines are:

17.5 After-Tax Evaluation 457
One project. PW or AW 0, the project is financially viable because the after-tax MARR is
met or exceeded.
Two or more alternatives. Select the alternative with the best (numerically largest) PW or AW
value.
If only cost CFAT amounts are estimated, calculate the after-tax savings generated by the operating
expenses and depreciation. Assign a plus sign to each saving and apply the guidelines above.
Remember, the equal-service assumption requires that the PW analysis be performed over the
least common multiple (LCM) of alternative lives. This requirement must be met for every
analysis—before or after taxes.
Since the CFAT estimates usually vary from year to year in an after-tax evaluation, the spreadsheet
offers a much speedier analysis than solution by hand.
For AW analysis: Use the PMT function with an embedded NPV function over one life cycle .
The general format is as follows, with the NPV function in italics for the CFAT series.
ME alternative selection
Equal service
PMT(MARR,n, NPV(MARR,year_1: year_n) year_0) [17.17]
For PW analysis: Obtain the PMT function results first, followed by the PV function taken
over the LCM. (There is an LCM function in Excel.) The cell containing the PMT function result
is entered as the A value. The general format is
EXAMPLE 17.6
PV(MARR,LCM_years, PMT_result_cell ) [17.18]
Paul is designing the interior walls of an industrial building. In some places, it is important
to reduce noise transmission across the wall. Two construction options—stucco on metal
lath (S) and bricks (B)—each have about the same transmission loss, approximately
33 decibels. This will reduce noise attenuation costs in adjacent office areas. Paul has estimated
the first costs and after-tax savings each year for both designs. (a) Use the CFAT
values and an after-tax MARR of 7% per year to determine which is economically better.
(b) Use a spreadsheet to select the alternative and determine the required first cost for the
plans to break even.
Plan S
Plan B
Year CFAT, $ Year CFAT, $
0 28,800 0 50,000
1–6 5,400 1 14,200
7–10 2,040 2 13,300
10 2,792 3 12,400
4 11,500
5 10,600
Solution by Hand
(a) In this example, both AW and PW analyses are shown. Develop the AW relations using
the CFAT values over each plan’s life. Select the larger value.
AW S [−28,800 5400(PA,7%,6) 2040(PA,7%,4)(PF,7%,6)
2792(PF,7%,10)](AP,7%,10)
$422
AW B [−50,000 14,200(PF,7%,1) · · · 10,600(PF,7%,5)](AP,7%,5)
$327
Both plans are financially viable; select plan S because AW S is larger.
For the PW analysis, the LCM is 10 years. Use the AW values and the PA factor for the
LCM of 10 years to select stucco on metal lath, plan S.
PW S AW S (PA,7%,10) 422(7.0236) $2964
PW B AW B (PA,7%,10) 327(7.0236) $2297

458 Chapter 17 After-Tax Economic Analysis
PMT(7%,5,NPV(7%,C4:C8)+C3)
PV(7%,10,$C$14)
(a)
Figure 17–5
(a) After-tax AW and PW analysis, and (b) breakeven first cost using Goal Seek, Example 17.6.
Solution by Spreadsheet
(b) Figure 17–5a, row 14, displays the AW value calculated using the PMT function defined by
Equation [17.17], and row 15 shows the 10-year PW that results from the PV function in
Equation [17.18]. Plan S is chosen by a relatively small margin.
Figure 17–5b shows the Goal Seek template used to equate the AW values and determine
the plan B first cost of $49,611 that causes the plans to break even. This is a small
reduction from the $−50,000 first cost initially estimated.
Comment
It is important to remember the minus signs in PMT and PV functions when utilizing them to
obtain the corresponding PW and AW values. If the minus is omitted, the AW and PW values
have the wrong sign and it appears that the plans are not financially viable in that they do not
return at least the after-tax MARR. That would happen in this example.
(b)
To utilize the ROR method, apply exactly the same procedures as in Chapter 7 (single project)
and Chapter 8 (two or more alternatives) to the CFAT series. A PW or AW relation is developed
to estimate the rate of return i * for a project, or i * for the incremental CFAT between two
alternatives. Multiple roots may exist in the CFAT series, as they can for any cash flow series. For
a single project, set the PW or AW equal to zero and solve for i *.
Project evaluation
Present worth:
Annual worth:
tn
0 CFAT t (PF,i*,t) [17.19]
t1
tn
0 CFAT t (PF,i*,t)(AP,i*,n) [17.20]
t1
If i * after-tax MARR, the project is economically justified.

17.5 After-Tax Evaluation 459
Spreadsheet solution for i * is faster for most CFAT series. It is performed using the IRR function
with the general format
IRR(year_0_CFAT:year_n_CFAT) [17.21]
If the after-tax ROR is important to the analysis, but the details of an after-tax study are not of
interest, the before-tax ROR (or MARR) can be adjusted with the effective tax rate T e by using
the approximating relation
Before-tax ROR ———————
after-tax ROR
1 T e
[17.22]
For example, assume a company has an effective tax rate of 40% and normally uses an after-tax
MARR of 12% per year for economic analyses that consider taxes explicitly. To approximate the
effect of taxes without performing the details of an after-tax study, the before-tax MARR can be
estimated as
Before-tax MARR ———— 0.12
20% per year
1 0.40
If the decision concerns the economic viability of a project and the resulting PW or AW value is
close to zero, the details of an after-tax analysis should be developed.
EXAMPLE 17.7
A fiber optics manufacturing company operating in Hong Kong has spent $50,000 for a
5- year-life machine that has a projected $20,000 annual NOI and annual depreciation of
$10,000 for years 1 through 5. The company has a T e of 40%. (a) Determine the after-tax rate
of return. (b) Approximate the before-tax return.
Solution
(a) The CFAT in year 0 is $50,000. For years 1 through 5 there is no capital purchase or sale,
so NOI CFBT. (See Equations [17.1] and [17.9].) Determine CFAT.
TI NOI D 20,000 10,000 $10,000
Taxes T e (TI) 0.4(10,000) $4000
CFAT CFBT taxes 20,000 4000 $16,000
Since the CFAT for years 1 through 5 has the same value, use the PA factor in Equation
[17.19].
0 50,000 16,000(PA,i*,5)
(PA,i*,5) 3.125
Solution gives i* 18.03% as the after-tax rate of return.
(b) Use Equation [17.22] for the before-tax return estimate.
Before-tax ROR ———— 0.1803 0.3005 (30.05%)
1 0.40
The actual before-tax i* using CFBT $20,000 for 5 years is 28.65% from the relation
0 50,000 20,000(PA,i*,5)
The tax effect will be slightly overestimated if a MARR of 30.05% is used in a before-tax
analysis.
A rate of return evaluation performed by hand on two or more alternatives utilizes a PW or
AW relation to determine the incremental return i * of the incremental CFAT series between
two alternatives. Solution by spreadsheet is accomplished using the incremental CFAT values
and the IRR function. The equations and procedures applied are the same as in Chapter 8
(Sections 8.4 through 8.6) for selection from mutually exclusive alternatives using the ROR
method. You should review and understand these sections before proceeding with this section.

460 Chapter 17 After-Tax Economic Analysis
From this review, several important facts can be recalled:
Selection guideline: The fundamental rule of incremental ROR evaluation at a stated MARR
is as follows:
Select the one alternative that requires the largest initial investment, provided the extra investment
is justified relative to another justified alternative.
Incremental ROR: Incremental analysis must be performed. Overall i * values cannot be
depended upon to select the correct alternative, unlike the PW or AW method at the MARR,
which will always indicate the correct alternative.
Equal-service requirement: Incremental ROR analysis requires that the alternatives be evaluated
over equal time periods. The LCM of the two alternative lives must be used to find the
PW or AW of incremental cash flows. (The only exception, mentioned in Section 8.5, occurs
when the AW analysis is performed on actual cash fl ows, not the increments; then onelife-cycle
analysis is acceptable over the respective alternative lives.)
Revenue and cost alternatives: Revenue alternatives (positive and negative cash flows) may
be treated differently from cost alternatives (cost-only cash flow estimates). For revenue alternatives,
the overall i * may be used to perform an initial screening. Alternatives with
i * MARR can be removed from further evaluation. An i * for cost-only alternatives cannot
be determined, so incremental analysis is required with all alternatives included.
Breakeven ROR
Once the CFAT series are developed, the breakeven ROR can be obtained using a plot of PW
versus i * by solving the PW relation for each alternative over the LCM at several interest rates.
For any after-tax MARR greater than the breakeven ROR, the extra investment is not justified.
The next examples solve CFAT problems using incremental ROR analysis and the breakeven
ROR plot of PW versus i .
EXAMPLE 17.8
In Example 17.6, Paul estimated the CFAT for interior wall materials to reduce sound transmission;
plan S is to construct with stucco on metal lath, and plan B is to construct using brick.
Figure 17–5a presented both a PW analysis over 10 years and an AW analysis over the respective
lives. Plan S was selected. After reviewing this earlier solution, (a) perform an ROR
evaluation at the after-tax MARR of 7% per year and (b) plot the PW versus i graph to determine
the breakeven ROR.
Solution by Spreadsheet
(a) The LCM is 10 years for the incremental ROR analysis, and plan B requires the extra investment
that must be justified. Apply the procedure in Section 8.6 for incremental ROR
analysis. Figure 17–6 shows the estimated CFAT for each alternative and the incremental
CFAT series. Since these are revenue alternatives, the overall i* is calculated first to ensure
that they both make at least the MARR of 7%. Row 14 indicates they do. The IRR
function (cell E14) is applied to the incremental CFAT, indicating that i* 6.35%. Since
this is lower than the MARR, the extra investment in brick walls is not justified. Plan S is
selected, the same as with the PW and AW methods.
(b) The NPV function is used to find the PW of the incremental CFAT series at various i values.
The graph indicates that the breakeven i* occurs at 6.35%—the same that the IRR function
found. Whenever the after-tax MARR is above 6.35%, as is the case here with MARR
7%, the extra investment in plan B is not justified.
Comment
Note that the incremental CFAT series has three sign changes. The cumulative series also has
three sign changes (Norstrom’s criterion). Accordingly, there may be multiple i* values. The
application of the IRR function using the “guess” option finds no other real number roots in the
normal rate of return range.

17.5 After-Tax Evaluation 461
Breakeven ROR
6.35%
Figure 17–6
Incremental evaluation of CFAT and determination of breakeven ROR, Example 17.8.
EXAMPLE 17.9
In Example 17.5 an after-tax analysis of two bone cell analyzers was initiated due to a new
3-year NBA contract. The criterion used to select analyzer 1 was the total taxes for the 3 years.
The complete solution is in Table 17–4 (hand) and Figure 17–4 (spreadsheet).
Continue the spreadsheet analysis by performing an after-tax ROR evaluation, assuming the
analyzers are sold after 3 years for the amounts estimated in Example 17.5: $130,000 for analyzer
1 and $225,000 for analyzer 2. The after-tax MARR is 10% per year.
Solution
A spreadsheet solution is presented here, but a hand solution is equivalent, just slower.
Fig ure 17–7 is an updated version of the spreadsheet in Figure 17–4 to include the sale of the
analyzers in year 3. The CFAT series (column I) are determined by the relation CFAT CFBT
taxes, with the taxable income determined using Equation [17.16], where DR is included. For
example, in year 3 when analyzer 2 is sold for S $225,000, the CFAT calculation is
CFAT 3 CFBT (TI)(T e ) GI OE P S (GI OE D DR)(T e )
The depreciation recapture DR is the amount above the year 3 book value received at sale time.
Using the book value after 3 years (F14),
DR selling price BV 3 225,000 64,800 $160,200
Now the CFAT in year 3 for analyzer 2 can be determined.
CFAT 3 100,000 10,000 0 225,000
(100,000 10,000 43,200 160,200)(0.35)
315,000 207,000(0.35) $242,550
The cell tags in row 14 of Figure 17–7 follow this same progression. The incremental CFAT is
calculated in column J, ready for the after-tax incremental ROR analysis.
These are revenue alternatives, so the overall i* values indicate that both CFAT series are
acceptable. The value i* 23.6% (cell J17) also exceeds MARR 10%, so analyzer 2 is
selected. This decision applies the ROR method guideline: Select the alternative that requires
the largest, incrementally justified investment.

462 Chapter 17 After-Tax Economic Analysis
Overall i *
B14C14E14(D14F14)
This term
calculates
DR $160,200
CFAT calculation
B14C14D14H14
Incremental i *
IRR(J11:J14)
Figure 17–7
Incremental ROR analysis of CFAT with depreciation recapture, Example 17.9.
Comment
In Section 8.4, we demonstrated the fallacy of selecting an alternative based solely on the
overall i*, because of the ranking inconsistency problem of the ROR method. The incremental
ROR must be used. The same fact is demonstrated in this example. If the larger i* alternative
is chosen, analyzer 1 is incorrectly selected. When i* exceeds the MARR, the larger investment
is correctly chosen—analyzer 2 in this case. For verification, the PW at 10% is calculated
for each analyzer (column I). Again, analyzer 2 is the winner, based on its larger PW of $93,905.
EXAMPLE 17.10
17.6 After-Tax Replacement Study
When a currently installed asset (the defender) is challenged with possible replacement, the effect
of taxes can have an impact upon the decision of the replacement study. The final decision
may not be reversed by taxes, but the difference between before-tax AW values may be significantly
different from the after-tax difference. Tax considerations in the year of the replacement
are as follows:
Depreciation recapture or tax savings due to a sizable capital loss are possible, if it is necessary
to trade the defender at a sacrifice price. Additionally, the after-tax replacement study considers
tax-deductible depreciation and operating expenses not accounted for in a before-tax analysis.
The effective tax rate T e is used to estimate the amount of annual taxes (or tax savings) from TI.
The same procedure as the before-tax replacement study in Chapter 11 is applied here, but for
CFAT estimates. The procedure should be thoroughly understood before proceeding. Special attention
to Sections 11.3 and 11.5 is recommended.
Example 17.10 presents a solution by hand of an after-tax replacement study using a simplifying
assumption of classical SL (straight line) depreciation. Example 17.11 solves the same problem
by spreadsheet, but includes the detail of MACRS depreciation. This provides an opportunity
to observe the difference in the AW values between the two depreciation methods.
Midcontinent Power Authority purchased emission control equipment 3 years ago for $600,000.
Management has discovered that it is technologically and legally outdated now. New equipment
has been identified. If a market value of $400,000 is offered as the trade-in for the current
equipment, perform a replacement study using (a) a before-tax MARR of 10% per year and

17.6 After-Tax Replacement Study 463
(b) a 7% per year after-tax MARR. Assume an effective tax rate of 34%. As a simplifying assumption,
use classical straight line depreciation with S 0 for both alternatives.
Defender
Challenger
Market value, $ 400,000
First cost, $
1,000,000
Annual cost, $year 100,000 15,000
Recovery period, years 8 (originally) 5
Solution
Assume that an ESL (economic service life) analysis has determined the best life values to be
5 more years for the defender and 5 years total for the challenger.
(a) For the before-tax replacement study, find the AW values. The defender AW uses the market
value as the first cost, P D $400,000.
AW D 400,000(AP,10%,5) 100,000 $205,520
AW C 1,000,000(AP,10%,5) 15,000 $278,800
Applying step 1 of the replacement study procedure (Section 11.3), we select the better AW
value. The defender is retained now with a plan to keep it for the 5 remaining years. The
defender has a $73,280 lower equivalent annual cost compared to the challenger. This complete
solution is included in Table 17–5 (left half) for comparison with the after-tax study.
(b) For the after-tax replacement study, there are no tax effects other than income tax for the
defender. The annual SL depreciation is $75,000, determined when the equipment was
purchased 3 years ago.
D t 600,0008 $75,000
t 1 to 8 years
Table 17–5 shows the TI and taxes at 34%. The taxes are actually tax savings of $59,500
per year, as indicated by the minus sign. (Remember that for tax savings in an economic
TABLE 17–5 Before-Tax and After-Tax Replacement Analyses, Example 17.10
Defender
Age
Year
Before Taxes
Expenses
OE, $ P and S, $ CFBT, $
Defender
Depreciation
D, $
Taxable
Income
TI, $
After Taxes
Taxes*
at
0.34TI, $ CFAT, $
3 0 400,000 400,000 400,000
4 1 100,000 100,000 75,000 175,000 59,500 40,500
5 2 100,000 100,000 75,000 175,000 59,500 40,500
6 3 100,000 100,000 75,000 175,000 59,500 40,500
7 4 100,000 100,000 75,000 175,000 59,500 40,500
8 5 100,000 0 100,000 75,000 175,000 59,500 40,500
AW at 10% 205,520 AW at 7% 138,056
Challenger
0 1,000,000 1,000,000 25,000 † 8,500 1,008,500
1 15,000 15,000 200,000 215,000 73,100 58,100
2 15,000 15,000 200,000 215,000 73,100 58,100
3 15,000 15,000 200,000 215,000 73,100 58,100
4 15,000 15,000 200,000 −215,000 73,100 58,100
5 15,000 0 15,000 200,000 215,000 ‡ 73,100 58,100
AW at 10% 278,800 AW at 7% 187,863
*
Minus sign indicates a tax savings for the year.
†
Depreciation recapture on defender trade-in.
‡
Assumes challenger’s salvage actually realized is S 0; no tax.

464 Chapter 17 After-Tax Economic Analysis
analysis it is assumed that there is positive taxable income elsewhere in the corporation to
offset the saving.) Since only costs are estimated, the annual CFAT is negative, but the
$59,500 tax savings has reduced it. The CFAT and AW at 7% per year are
CFAT CFBT taxes 100,000 (59,500) $40,500
AW D 400,000(AP,7%,5) 40,500 $138,056
For the challenger, depreciation recapture on the defender occurs when it is replaced
because the trade-in amount of $400,000 is larger than the current book value. In year
0 for the challenger, Table 17–5 includes the following computations to arrive at a tax
of $8500.
Defender book value, year 3: BV 3 600,000 3(75,000) $375,000
Depreciation recapture: DR 3 TI 400,000 375,000 $25,000
Taxes on trade-in, year 0: Taxes 0.34(25,000) $8500
The SL depreciation is $1,000,0005 $200,000 per year. This results in tax saving and
CFAT as follows:
Taxes (15,000 200,000)(0.34) $73,100
CFAT CFBT taxes 15,000 (73,100) $58,100
In year 5, it is assumed the challenger is sold for $0; there is no depreciation recapture. The
AW for the challenger at the 7% after-tax MARR is
AW C 1,008,500(AP,7%,5) 58,100 $187,863
The defender is again selected; however, the equivalent annual advantage has decreased
from $73,280 before taxes to $49,807 after taxes.
Conclusion: By either analysis, retain the defender now and plan to keep it for 5 more
years. Additionally, plan to evaluate the estimates for both alternatives 1 year hence. If and
when cash flow estimates change significantly, perform another replacement analysis.
Comment
If the market value (trade-in) had been less than the current defender book value of $375,000, a
capital loss, rather than depreciation recapture, would occur in year 0. The resulting tax savings
would decrease the CFAT (which is to reduce costs if CFAT is negative) of the challenger. For
example, a trade-in amount of $350,000 would result in a TI of $350,000 375,000 $25,000
and a tax savings of $8500 in year 0. The CFAT is then $1,000,000 (8500)
$991,500.
EXAMPLE 17.11
Repeat the after-tax replacement study of Example 17.10b using 7-year MACRS depreciation
for the defender and 5-year MACRS depreciation for the challenger. Assume either
asset is sold after 5 years for exactly its book value. Determine if the answers are significantly
different from those obtained when the simplifying assumption of classical SL depreciation
was made.
Solution
Figure 17–8 shows the complete analysis. MACRS requires more computation than SL depreciation,
but this effort is easily reduced by the use of a spreadsheet. Again the defender is selected
for retention, but now by an advantage of $44,142 annually. This compares to the
$49,807 advantage using classical SL depreciation and the $73,280 before-tax advantage of the
defender. Therefore, taxes and MACRS have reduced the defender’s economic advantage, but
not enough to reverse the decision to retain it.
Several other differences in the results between SL and MACRS depreciation are worth
noting. There is depreciation recapture in year 0 of the challenger due to trade-in of the defender
at $400,000, a value larger than the book value of the 3-year-old defender. This amount,

17.7 After-Tax Value-Added Analysis 465
Depreciation recapture
C5 F11
Challenger sales price
B15 F24
Figure 17–8
After-tax replacement study with MACRS depreciation and depreciation recapture, Example 17.11.
$137,620 (cell G18), is treated as ordinary taxable income. The calculations for the DR and
associated tax, by hand, are as follows:
BV 3 first cost MACRS depreciation for 3 years
total MACRS depreciation for years 4 through 8
$262,380 (cell F11)
DR TI 0 trade-in BV 3
400,000 − 262,380 $137,620 (cell G18)
Taxes (0.34)(137,620) $46,791 (cell H18)
See the cell tags and table notes that duplicate this logic.
The assumption that the challenger is sold after 5 years at its book value implies a positive
cash flow in year 5. The entry $57,600 (C23) reflects this assumption, since the forgone
MACRS depreciation in year 6 would be 1,000,000(0.0576) $57,600. The spreadsheet
relation B15 F24 determines this value using the accumulated depreciation in F24.
[Note: If the salvage S 0 is anticipated after 5 years, then a capital loss of $57,600 will be
incurred. This implies an additional tax saving of 57,600(0.34) $19,584 in year 5. Conversely,
if the salvage value exceeds the book value, a depreciation recapture and associated
tax should be estimated.]
17.7 After-Tax Value-Added Analysis
When a person or company is willing to pay more for an item, it is likely that some processing
has been performed on an earlier version of the item to make it more valuable now to the purchaser.
This is value added.
Value added is a term used to indicate that a product or service has added worth from the
perspective of a consumer, owner, investor, or purchaser. It is common to leverage valueadding
activities on a product or service.
Value added

466 Chapter 17 After-Tax Economic Analysis
For an example of highly leveraged value-added activities, consider onions that are grown and
sold at the farm level for cents per pound. They may be purchased by the shopper in a store at
50 cents to $1.25 per pound. But when onions are cut and coated with a special batter, they may
be fried in hot oil and sold as onion rings for several dollars per pound. Thus, from the perspective
of the consumer, there has been a large amount of value added by the processing from raw
onions in the ground into onion rings sold at a restaurant or fast-food shop.
The value-added measure was briefly introduced in conjunction with AW analysis before
taxes. When value-added analysis is performed after taxes, the approach is somewhat different
from that of CFAT analysis developed previously in this chapter. However, as shown below,
The decision about an alternative will be the same for both the value-added and CFAT methods,
because the AW of economic value-added estimates is the same as the AW of CFAT estimates.
Value-added analysis starts with Equation [17.4], net operating profit after taxes (NOPAT), which
includes the depreciation for year 1 through year n. Depreciation D is included in that TI GI
OE − D . This is different from CFAT, where the depreciation has been specifically removed so that
only actual cash flow estimates are used for year 0 through year n .
The term economic value added (EVA) indicates the monetary worth added by an alternative
to the corporation’s bottom line. (The term EVA is a trademark of Stern Stewart & Co.) The technique
discussed below was first publicized in several articles 1 in the mid-1990s, and it has since
become very popular as a means to evaluate the ability of a corporation to increase its economic
worth, especially from the shareholders’ viewpoint.
The annual EVA is the amount of NOPAT remaining on corporate books after removing the cost
of invested capital during the year. That is, EVA indicates the project’s contribution to the net
profit of the corporation after taxes.
The cost of invested capital is the after-tax rate of return (usually the MARR value) multiplied
by the book value of the asset during the year. This is the interest incurred by the current
level of capital invested in the asset. (If different tax and book depreciation methods are used, the
book depreciation value is used here, because it more closely represents the remaining capital
invested in the asset from the corporation’s perspective.) Computationally,
EVA NOPAT cost of invested capital
NOPAT (after-tax interest rate)(book value in year t 1)
TI(1 T e ) ( i )(BV t 1 ) [17.23]
Since both TI and the book value consider depreciation, EVA is a measure of worth that mingles
actual cash flow with noncash flows to determine the estimated financial worth contribution to
the corporation. This financial worth is the amount used in public documents of the corporation
(balance sheet, income statement, stock reports, etc.). Because corporations want to present the
largest value possible to the stockholders and other owners, the EVA method is often more appealing
than the AW method from the financial perspective.
The result of an EVA analysis is a series of annual EVA estimates. Two or more alternatives
are compared by calculating the AW of EVA estimates and selecting the alternative with the
larger AW value. If only one project is evaluated, AW 0 means the after-tax MARR is exceeded,
thus making the project value-adding.
Sullivan and Needy 2 have demonstrated that the AW of EVA and the AW of CFAT are identical
in amount. Thus, either method can be used to make a decision. The annual EVA estimates indicate
added worth to the corporation generated by the alternative, while the annual CFAT estimates
describe how cash will flow. This comparison is made in Example 17.12 .
1 A. Blair, “EVA Fever,” Management Today, Jan. 1997, pp. 42–45; W. Freedman, “How Do You Add Up?”
Chemical Week, Oct. 9, 1996, pp. 31–34.
2 W. G. Sullivan and K. L. Needy, “Determination of Economic Value Added for a Proposed Investment in
New Manufacturing.” The Engineering Economist, vol. 45, no. 2 (2000), pp. 166–181.

17.7 After-Tax Value-Added Analysis 467
EXAMPLE 17.12
Biotechnics Engineering has developed two mutually exclusive plans for investing in new
capital equipment with the expectation of increased revenue from its medical diagnostic services
to cancer patients. The estimates are summarized below. (a) Use classical straight line
depreciation, an after-tax MARR of 12%, and an effective tax rate of 40% to perform two annual
worth after-tax analyses: EVA and CFAT. (b) Explain the fundamental difference between
the results of the two analyses.
Plan A
Plan B
Initial investment, $ 500,000 1,200,000
Gross income expenses, $ 170,000 per year 600,000 in year 1, decreasing by 100,000
per year thereafter
Estimated life, years 4 4
Salvage value None None
Solution by Spreadsheet
(a) Refer to the spreadsheet and function cells (row 22) in Figure 17–9.
EVA evaluation: All the necessary information for EVA estimation is determined in
columns B through G. The net operating profit after taxes (NOPAT) in column H is calculated
by Equation [17.4], TI taxes. The book values (column E) are used to determine the
cost of invested capital in column I, using the second term in Equa tion [17.23], that is,
i(BV t1 ), where i is the 12% after-tax MARR. This represents the amount of interest at
12% per year, after taxes, for the currently invested capital as reflected by the book value
at the beginning of the year. The EVA estimate is the sum of columns H and I for years 1
through 4. Notice there is no EVA estimate for year 0, since NOPAT and the cost of invested
capital are estimated for years 1 through n. Finally, the larger AW of the EVA value is selected,
which indicates that plan B is better and that plan A does not make the 12% return.
CFAT evaluation: As shown in function row 22 (plan B for year 3), CFAT estimates
(column K) are calculated as (GI OE) P taxes. The AW of CFAT again concludes
that plan B is better and that plan A does not return the after-tax MARR of 12% (K10).
(b) What is the fundamental difference between the EVA and CFAT series in columns J and K?
They are clearly equivalent from the time value of money perspective since the AW values
are numerically the same. To answer the question, consider plan A, which has a constant
CFAT estimate of $152,000 per year. To obtain the AW of EVA estimate of $12,617 for
Figure 17–9
Comparison of two plans using EVA and CFAT analyses, Example 17.12.

468 Chapter 17 After-Tax Economic Analysis
years 1 through 4, the initial investment of $500,000 is distributed over the 4-year life using
the AP factor at 12%. That is, an equivalent amount of $500,000(AP,12%,4) $164,617
is “charged” against the cash inflows in each of years 1 through 4. In effect, the yearly
CFAT is reduced by this charge.
CFAT (initial investment)(AP,12%,4) 152,000 − 500,000(AP,12%,4)
152,000 164,617 $12,617
AW of EVA
This is the AW value for both series, demonstrating that the two methods are economically
equivalent. However, the EVA method indicates an alternative’s yearly estimated contribution
to the value of the corporation, whereas the CFAT method estimates the actual cash
flows to the corporation. This is why the EVA method is often more popular than the cash
flow method with corporate executives.
Comment
The calculation P(AP,i,n) $500,000(AP,12%,4) is exactly the same as the capital recovery
in Equation [6.3], assuming an estimated salvage value of zero. Thus, the cost of invested
capital for EVA is the same as the capital recovery discussed in Chapter 6. This further demonstrates
why the AW method is economically equivalent to the EVA evaluation.
17.8 After-Tax Analysis for International Projects
Primary questions to be answered prior to performing a corporate-based after-tax analysis for
international settings revolve around tax-deductible allowances—depreciation, business expenses,
capital asset evaluation—and the effective tax rate needed for Equation [17.7], taxes
( T e )(TI). As discussed in Chapter 16, most governments of the world recognize and use the
straight line (SL) and declining balance (DB) methods of depreciation with some variations to
determine the annual tax-deductible allowance. Expense deductions vary widely from country to
country. By way of example, some of these are summarized here.
Canada
Depreciation: This is deductible and is normally based on DB calculations, although SL may be
used. An equivalent of the half-year convention is applied in the first year of ownership. The
annual tax-deductible allowance is termed capital cost allowance (CCA). As in the U.S. system,
recovery rates are standardized, so the depreciation amount does not necessarily reflect
the useful life of an asset.
Class and CCA rate: Asset classes are defined and annual depreciation rates are specified by
class. No specific recovery period (life) is identified, in part because assets of a particular class
are grouped together and the annual CCA is determined for the entire class, not individual assets.
There are some 44 classes, and CCA rates vary from 4% per year (the equivalent of a
25-year-life asset) for buildings (class 1) to 100% (1-year life) for applications software, chinaware,
dies, etc. (class 12). Most rates are in the range of 10% to 30% per year.
Expenses: Business expenses are deductible in calculating TI. Expenses related to capital investments
are not deductible, since they are accommodated through the CCA.
Internet: Further details are available on the Revenue Canada website at www.cra.gc.ca in the
Forms and Publications section.
China (PRC)
Depreciation: Officially, SL is the primary method of tax-deductible depreciation; however, assets
employed in selected industries or types of assets can utilize accelerated DB or SYD
(sum-of-years-digits) depreciation, when approved by the government. The selected industries
and assets can change over time; currently favored industries serve areas such as technology
and oil exploration, and equipment subjected to large vibrations during normal usage is
allowed accelerated depreciation.

17.8 After-Tax Analysis for International Projects 469
Recovery period: Standardized recovery periods are published that vary from 3 years (electronic
equipment) to 10 years (aircraft, machinery, and other production equipment) to 20 years (buildings).
Shortened periods can be approved, but the minimum recovery period cannot be less
than 60% for the normal period defined by current tax law.
Expenses: Business expenses are deductible with some limitations and some special incentives.
Limitations are placed, for example, on advertising expense deductions (15% of sales for the
year). Incentives are generous in some cases; for example, 150% of actual expenses is deductible
for new technology and new product R&D activities.
Internet: Summary information for China and several other countries is available at
www.worldwide-tax.com.
Mexico
Depreciation: This is a fully deductible allowance for calculating TI. The SL method is applied
with an index for inflation considered each year. For some asset types, an immediate deduction
of a percentage of the first cost is allowed. (This is a close equivalent to the Section 179
Deduction in the United States.)
Class and rates: Asset types are identified, though not as specifically defined as in some countries.
Major classes are identified, and annual recovery rates vary from 5% for buildings (the
equivalent of a 20-year life) to 100% for environmental machinery. Most rates range from
10% to 30% per year.
Profi t tax: The income tax is levied on profits on income earned from carrying on business in
Mexico. Most business expenses are deductible. Corporate income is taxed only once, at the
federal level; no state-level taxes are imposed.
Tax on Net Assets (TNA): Under some conditions, a tax of 1.8% of the average value of assets
located in Mexico is paid annually in addition to income taxes.
Internet: The best information is via websites for companies that assist international corporations
located in Mexico. One example is PriceWaterhouseCoopers at www.pwcglobal
.com/mx/eng.
Japan
Depreciation: This is fully deductible and based on classical SL and DB methods. To encourage
capital investment for long-term economic growth, in 2007 Japan allowed assets to be depreciated
using a 250% DB method ; that is, a significantly increased accelerated rate is allowed
compared to historical Japanese allowances. Switching to classical SL depreciation must take
place in the year that the accelerated rate amount falls below the corresponding SL amount.
Class and life: A statutory useful life ranging from 4 to 24 years, with a 50-year life for reinforced
concrete buildings, is specified.
Expenses: Business expenses are deductible in calculating TI.
Internet: Further details are available on the Japanese Ministry of Finance website at
www.mof.go.jp.
The effective tax rate varies considerably among countries. Some countries levy taxes only
at the federal level, while others impose taxes at several levels of government (federal, state or
provincial, prefecture, county, and city). A summary of international corporate average tax rates
is presented in Table 17–6 for a wide range of industrialized countries. These include income
taxes at all reported levels of government within each country; however, other types of taxes
may be imposed by a particular government. Although these average rates of taxation will vary
from year to year, especially as tax reform is enacted, it can be surmised that most corporations
face effective rates of about 20% to 40% of taxable income. A close examination of international
rates shows that they have decreased significantly over the last decade. In fact, the KPMG
report noted in Table 17–6 indicates that the global average corporate tax rate on TI has decreased
from 32.7% (1999) to 25.5% (2009). This has encouraged corporate investment and
business expansion within country borders and helped soften the massive economic downturn
experienced in recent years.

470 Chapter 17 After-Tax Economic Analysis
TABLE 17–6
Summary of International Corporate Average Tax Rates
Tax Rate Levied on
Taxable Income, %
For These Countries
40
United States, Japan
35 to 40 South Africa, Pakistan
32 to 35 Canada, France, India
28 to 32 Australia, United Kingdom, New Zealand, Spain, Germany,
Mexico, Indonesia
24 to 28 China, Taiwan, Republic of Korea
20 to 24 Russia, Turkey, Saudi Arabia
20
Singapore, Hong Kong, Chile, Ireland, Iceland, Hungary
Sources: Extracted from KPMG’s Corporate and Indirect Tax Survey 2009 (www.kpmg.
comGlobalenIssuesAndInsightsPagesdefault.aspx) and from country websites on corporate taxation.
One of the prime ways that governments have been able to reduce corporate tax rates is by
shifting to indirect taxes on goods and services for additional tax revenue. These taxes are usually
in the form of a value-added tax (VAT), goods and services tax (GST), and taxes on products
imported from outside its borders. As corporate tax rates have declined, in general the indirect
tax rates have increased. This has been especially true during the first decade of the 21st
century. However, the worldwide economic slump experienced in recent years has made it necessary
for governments around the world to be more cautious of how they tax corporations and
maintain a reasonable balance between regular tax rates (as listed in Table 17–6 ) and indirect tax
rates. The VAT system is explained now.
17.9 Value-Added Tax
A value-added tax (VAT) has facetiously been called a sales tax on steroids, because the VAT
rates on some items in some countries that impose a VAT can be as high as 90%. It is also called
a GST (goods and service tax).
A value-added tax is an indirect tax; that is, it is a tax on goods and services rather than on
people or corporations. It differs from a sales tax in two ways: (1) when it is charged and (2) who
pays (as explained below). A specific percentage, say 10%, is a charge added to the price of the
item and paid by the buyer. The seller then sends this 10% VAT to the taxing entity, usually a
government unit. This process of 10% VAT continues every time the item is resold—as purchased
or in a modified form—thus the term value-added .
A value-added tax is commonly used throughout the world. In some countries, VAT is used in
lieu of business or individual income taxes. There is no VAT system in the United States yet. In
fact, the United States is the only major industrialized country in the world that does not have a
VAT system, though other forms of indirect taxation are used liberally. There is mounting evidence,
however, that a VATGST system will be necessary in the United States in the near future,
but not without a great amount of political discord.
A sales tax is used by the U.S. government, by nearly all of the states, and by many local entities.
A sales tax is charged on goods and services at the time the goods and services reach the end
user or consumer . That is, businesses do not pay a sales tax on raw material, unfinished goods,
or items they purchase that will ultimately be sold to an end user; only the end user pays the sales
tax. Businesses do pay a sales tax on items for which they are the end user. Total sales tax percentages
imposed by multiple government levels can range from 5% to 11%, sometimes larger
on specific items. For example, when Home Depot (HD) purchases microwave ovens from Jenn-
Air, Inc., Home Depot does not pay a sales tax on the microwave ovens because they will be sold
to HD customers who will pay the sales tax. On the other hand, if Home Depot purchases a forklift
from Caterpillar for loading and unloading merchandise in one of its stores, HD will pay the
sales tax on the forklift, since HD is the end user. Thus, a sales tax is paid only one time , and that

17.9 Value-Added Tax 471
time is when the goods or services are purchased by the end user . The sales tax is the responsibility
of the merchant to collect and remand to the taxing entity.
A value-added tax (VAT) , on the other hand, is charged to the buyer at purchase time, whether
the buyer is a business or end user. The seller sends the collected VAT to the taxing entity. If the
buyer subsequently resells the goods to another buyer (as is or a modification of it), another VAT
is collected by the seller. Now, this second seller will send to the taxing entity an amount that is
equal to the total tax collected minus the amount of VAT already paid.
As an illustration, assume the U.S. government charged a 10% VAT. Here is how the VAT
might work.
Northshore Mining Corporation of Babbitt, Minnesota, sells $100,000 worth of iron ore to
Westfall Steel. As part of the price, Northshore collects $110,000, that is, $100,000 0.1
100,000, from Westfall Steel. Northshore remits the $10,000 VAT to the U.S. Treasury.
Westfall Steel sells all of the steel it made from the iron ore at a price of $300,000 to General
Electric (GE). Westfall collects $330,000 from GE and then sends $20,000 to the U.S. Treasury,
that is, $30,000 it collected in VAT from GE minus $10,000 it paid in taxes to Northshore
Mining.
GE uses the steel to make refrigerators that it sells for $700,000 to retailers, such as Home
Depot, Lowe’s, and others. GE collects $770,000 and then remits $40,000 to the U.S. Treasury,
that is, $70,000 it collected in taxes from retailers minus $30,000 it paid in VAT to
Westfall Steel.
If GE purchased machines, tools, or other items to make the refrigerators during this accounting
period, and it paid taxes on those items, the taxes paid would also be deducted from the
VAT that GE collected before sending the money to the U.S. Treasury. For example, if GE
paid $5000 in taxes on motors it purchased for the refrigerators, the amount GE would remit
to the U.S. Treasury would be $35,000 (that is, $70,000 it collected from retailers minus
$30,000 it paid in taxes to Westfall Steel minus $5000 it paid in taxes on the motors).
The retailers sell the refrigerators for $950,000 and collect $95,000 in taxes from end users—
consumers. The retailers remit $25,000 to the U.S. Treasury (that is, $95,000 they collected
minus $70,000 they paid previously).
Through this process, the U.S Treasury has received $10,000 from Northshore, $20,000
from Westfall Steel, $35,000 from GE, $5000 from the supplier of the motors, and $25,000
from the retailers, for a total of $95,000. This is 10% of the final sales price of $950,000. The
VAT money was deposited into the Treasury at several different times from several different
companies.
The taxes that a company pays for materials or items it purchases in order to produce goods
or services that will subsequently be sold to another business or end user are called input taxes ,
and the company is able to recover them when it collects the VAT from the sale of its products.
The taxes that a company collects are called output taxes , and these are forwarded to the taxing
entity, less the amount of input taxes the company paid. Hence, the businesses incur no taxes
themselves, the same as with a sales tax.
Several dimensions of a VAT distinguish it from a sales tax or corporate income taxes. Some
are as follow:
• Value-added taxes are taxes on consumption, not production or taxable income.
• The end user pays all of the value-added taxes, but VATs are not as obvious as a sales tax that
is added to the price of the item at the time of purchase (and displayed on the receipt). Therefore,
VAT taxing entities encounter less resistance from consumers.
• Value-added taxes are generally considerably higher than sales taxes, with the average
European VAT rate at 20% and the worldwide average at 15.25%.
• The VAT is essentially a “sales tax,” but it is charged at each stage of the product development
process instead of when the product is sold.
• With VAT, there is less evasion of taxes because it is harder for multiple entities to evade collecting
and paying the taxes than it is for one entity to do so.
• VAT rates vary from country to country and from category to category. For example, in some
countries, food has a 0% VAT rate, while aviation fuel is taxed at 32%.

472 Chapter 17 After-Tax Economic Analysis
EXAMPLE 17.13
Tata Motors is a major player in automobile manufacturing in India. It has three different
manufacturing units that specialize in manufacturing different transportation-related products,
such as trucks, engines and axles, commercial vehicles, utility vehicles, and passenger
cars. The company buys products that fall under different sections of the Indian government
VAT tax code, and therefore the products have different VAT rates. In one
particular accounting period, Tata had invoices from four different suppliers (vendors A,
B, C, and D) in the respective amounts of $1.5 million, $3.8 million, $1.1 million, and
$900,000. The products Tata purchased were subject to VAT rates of 4%, 4%, 12.5%, and
22%, respectively.
(a) How much total VAT did Tata pay to its vendors?
(b) Assume that Tata’s products have a VAT rate of 12.5%. If Tata’s sales during the period
were $9.2 million, how much VAT did the Indian Treasury receive from Tata?
Solution
(a) Let X equal the product price before the VAT is added. Solve for X and then subtract it
from the purchase amount to determine the VAT charged by each vendor. Table 17–7
shows the VAT that Tata paid its four vendors. An example computation for vendor A is
as follows:
X 0.04X 1,500,000
1.04X 1,500,000
X $1,442,308
VAT A 1,500,000 1,442,308
$57,692
Total VAT paid 57,692 146,154 122,222 162,295
$488,363
(b) Total from Tata total VAT VAT paid by vendors
9,200,000(0.125) 488,363
$661,637
TABLE 17–7 VAT Computation, Example 17.13
Vendor Purchases, $ VAT Rate, % Price before VAT, X, $ VAT, $
A 1,500,000 4.0 1,442,308 57,692
B 3,800,000 4.0 3,653,846 146,154
C 1,100,000 12.5 977,778 122,222
D 900,000 22.0 737,705 162,295
Total 488,363
CHAPTER SUMMARY
After-tax analysis does not usually change the decision to select one alternative over another;
however, it does offer a much clearer estimate of the monetary impact of taxes. After-tax PW,
AW, and ROR evaluations of one or more alternatives are performed on the CFAT series using
exactly the same procedures as in previous chapters.
Income tax rates for U.S. corporations and individual taxpayers are graduated or progressive—
higher taxable incomes pay higher income taxes. A single-value, effective tax rate T e is usually
applied in an after-tax economic analysis. Taxes are reduced because of tax deductible items,
such as depreciation and operating expenses. Because depreciation is a noncash flow, it is important
to consider depreciation only in TI computations, and not directly in the CFBT and

CFAT calculations. Accordingly, key general cash flow after-tax relations for each year are as
follows:
NOI gross income expenses
TI gross income operating expenses depreciation depreciation recapture
CFBT gross income operating expenses initial investment salvage value
CFAT CFBT taxes CFBT ( T e )(TI)
If an alternative’s estimated contribution to corporate financial worth is the economic measure,
the economic value added (EVA) should be determined. Unlike CFAT, the EVA includes the effect
of depreciation. The equivalent annual worths of CFAT and EVA estimates are the same
numerically, because they interpret the annual cost of the capital investment in different, but
equivalent manners when the time value of money is taken into account.
In a replacement study, the tax impact of depreciation recapture, which may occur when the
defender is traded for the challenger, is accounted for in an after-tax analysis. The replacement
study procedure of Chapter 11 is applied. The tax analysis may not reverse the decision to replace
or retain the defender, but the effect of taxes will likely reduce (possibly by a significant amount)
the economic advantage of one alternative over the other.
International corporate tax rates have steadily decreased, but indirect taxes, such as valueadded
tax (VAT), have increased. The mechanism of a VAT is explained and compared to a sales
tax. The United States currently has no VAT.
Problems 473
PROBLEMS
Basic Tax Computations
17.1 (a) Define the following tax terms: graduated tax
rates , marginal tax rate , and indexing .
( b) Describe a fundamental difference between
each of the following terms: net operating income
(NOI), taxable income (TI), and net
operating profit after taxes (NOPAT).
17.2 Determine the average tax rate for a corporation
that has taxable income of ( a ) $150,000 and
( b ) $12,000,000.
17.3 Determine the single-value effective tax rate for a
corporation that has a federal tax rate of 35% and a
state tax rate of 5%.
17.4 Identify the primary term described by each event
below: gross income , depreciation , operating expense
, taxable income , income tax , or net operating
profit after taxes .
( a) A new machine had a first-year write-off of
$10,500.
( b) A public corporation estimates that it will report
a $−750,000 net profit on its annual income
statement.
( c) An asset with a book value of $8000 was retired
and sold for $8450.
( d) An over-the-counter software system will
generate $420,000 in revenue this quarter.
( e) An asset with a MACRS recovery period of 7
years has been owned for 10 years. It was
just sold for $2750.
(f ) The cost of goods sold in the past year was
$3,680,200.
( g) A convenience store collected $33,550 in lottery
ticket sales last month. Based on winners
holding these tickets, a rebate of $350
was sent to the manager.
( h) An asset with a first cost of $65,000 was utilized
on a new product line to increase sales
by $150,000.
( i) The cost to maintain equipment during the
past year was $641,000.
17.5 Two companies have the following values on their
annual tax returns.
Company 1 Company 2
Sales revenue, $ 1,500,000 820,000
Interest revenue, $ 31,000 25,000
Expenses, $ 754,000 591,000
Depreciation, $ 48,000 18,000
(a) Calculate the federal income tax for the year
for each company.
(b) Determine the percent of sales revenue each
company will pay in federal income tax.
(c) Estimate the taxes using an effective rate of
34% of the entire TI. Determine the percentage
error made relative to the exact taxes in
part ( a ).
17.6 Last year, one division of Hagauer.com, a dot-com
sports industry service firm that provides real-time

474 Chapter 17 After-Tax Economic Analysis
analysis of mechanical stress due to athlete injury,
had $250,000 in taxable income. This year, TI is
estimated to be $600,000. Calculate the federal income
taxes and answer the following.
(a) What was the average federal tax rate paid
last year?
(b) What is the marginal federal tax rate on the
additional TI this year?
(c)
What will be the average federal tax rate this
year?
(d) What will be the NOPAT on just the additional
$350,000 in taxable income?
17.7 Yamachi and Nadler of Hawaii has a gross income
of $7.5 million for the year. Depreciation and operating
expenses total $4.3 million. The combined
state and local tax rate is 7.2%. If an effective federal
rate of 35% applies, estimate the income taxes
using the effective tax rate equation.
17.8 Workman Tools reported a TI of $90,000 last year.
If the state income tax rate is 7%, determine the
( a ) average federal tax rate, ( b ) overall effective
tax rate, ( c ) total taxes to be paid based on the effective
tax rate, and ( d ) total taxes paid to the state
and paid to the federal government.
17.9 The taxable income for a motorcycle sales and repair
business is estimated to be $150,000 this year.
A single-value tax rate of 39% is used. A new engine
diagnostics system will cost $40,000 and
have an average annual depreciation of $8000. The
equipment will increase gross income by an estimated
$9000 and expenses by $2000 for the year.
Compute the expected change in income taxes for
the year if the new system is purchased.
17.10 C.F. Jordon Construction Services has operated for
the last 26 years in a northern U.S. state where the
state income tax on corporate revenue is 6% per
year. C.F. Jordon pays an average federal tax of
23% and reports taxable income of $7 million. Because
of pressing labor cost increases, the president
wants to move to another state to reduce the
total tax burden. The new state may have to be
willing to offer tax allowances or an interest-free
grant for the first couple of years in order to attract
the company. You are an engineer with the company
and are asked to do the following.
(a) Determine the effective tax rate for C.F.
Jordon.
(b) Estimate the state tax rate that would be necessary
to reduce the overall effective tax rate
by 10% per year.
(c) Determine what the new state would have to
do financially for C.F. Jordon to move there
and to reduce its effective tax rate to 22% per
year.
CFBT and CFAT
17.11 Identify which of the following items are included
in the calculation of cash flow before taxes
(CFBT): operating expenses, salvage value, depreciation,
initial investment, gross income, tax rate.
17.12 What is the basic difference between cash flow
after taxes (CFAT) and net operating profit after
taxes (NOPAT)?
17.13 For a year in which there is no initial investment P
or salvage value S , derive an equation for CFBT
that contains only the following terms: CFAT,
CFBT, D , and T e .
17.14 Determine the cash flow before taxes for Anderson
Consultants when the cash flow after taxes was
$600,000, asset depreciation was $350,000 and the
company’s effective tax rate was 36%.
17.15 Four years ago Sierra Instruments of Monterey,
California spent $200,000 for equipment to manufacture
standard gas flow calibrators. The equipment
was depreciated by MACRS using a 3-year
recovery period. The gross income for year 4 was
$100,000, with operating expenses of $50,000.
Use an effective tax rate of 40% to determine the
CFAT in year 4 if the asset was ( a ) discarded with
no salvage value in year 4 and ( b ) sold for $20,000
at the end of year 4 (neglect any taxes that may be
incurred on the sale of the equipment). The
MACRS depreciation rate for year 4 is 7.41%.
17.16 Four years ago a division of Harcourt-Banks purchased
an asset that was depreciated by the MACRS
method using a 3-year recovery period. The total
revenue for year 2 was $48 million, depreciation
was $8.2 million, and operating expenses were
$28 million. Use a federal tax rate of 35% and a
state tax rate of 6.5% to determine ( a ) CFAT, ( b )
percentage of total revenue expended on taxes, and
( c ) net profit after taxes for the year.
17.17 Advanced Anatomists, Inc., researchers in medical
science, is contemplating a commercial venture
concentrating on proteins based on the new
X-ray technology of free- electron lasers. To recover
the huge investment needed, an annual
$2.5 million CFAT is needed. A favored average
federal tax rate of 20% is expected; however, state
taxing authorities will levy an 8% tax on TI. Over
a 3-year period, the deductible expenses and depreciation
are estimated to total $1.3 million the
first year, increasing by $500,000 per year thereafter.
Of this, 50% is expenses and 50% is depreciation.
What is the required gross income each
year?

Problems 475
The following information is used in Problems 17.18
through 17.21. (Show hand and spreadsheet solutions, as
instructed)
After 4 years of use, Procter and Gamble has decided to
replace capital equipment used on its Zest bath soap line.
The equipment was MACRS-depreciated over a 3-year
recovery period. After-tax MARR is 10% per year, and T e
is 35% in the United States. The cash flow data is tabulated
in $1000 units.
Year
0 1 2 3 4
Purchase, $
1900
Gross income, $ 800 950 600 300
Operating expenses, $ 100 150 200 250
Salvage, $ 700
17.18 Utilize the CFBT series and AW value to determine
whether the equipment investment exceeded
the MARR.
17.19 Calculate MACRS depreciation and estimate the
CFAT series over 4 years. Neglect any tax impact
caused by the $700,000 salvage received in year 4.
17.20 Utilize the CFAT series and AW value to determine
whether the investment exceeded the MARR.
17.21 Compare the after-tax ROR values using both
methods—CFAT series and approximation from
the CFBT values using the before-tax ROR and T e .
17.22 A Wal-Mart Distribution Center has put into service
forklifts and conveyors purchased for $250,000.
Use a spreadsheet to tabulate CFBT, CFAT, NOPAT,
and i * before and after taxes for 6 years of ownership.
The effective tax rate is 40%, and the estimated
cash flow and depreciation amounts are
shown. Salvage is expected to be zero.
Year
Gross
Income, $
Operating
Expenses, $
MACRS
Depreciation, $
1 210,000 120,000 50,000
2 210,000 120,000 80,000
3 160,000 122,000 48,000
4 160,000 124,000 28,800
5 160,000 126,000 28,800
6 140,000 128,000 14,400
Taxes and Depreciation
17.23 An asset purchased by Stratasys, Inc. had a first
cost of $70,000 with an expected salvage value of
$10,000 at the end of its 5-year life. In year 2, the
revenue was $490,000 with operating expenses of
$140,000. If the company’s effective tax rate was
36%, determine the difference in taxes paid in
year 2 if the depreciation method had been straight
line instead of MACRS.
17.24 Cheryl, an electrical engineering student who is
working on a business minor, is studying depreciation
and finance in her engineering management
course. The assignment is to demonstrate that
shorter recovery periods require the same total
taxes, but they offer a time value of taxes advantage
for depreciable assets. Help her, using asset
estimates made for a 6-year study period: P
$65,000, S $5000, GI $32,000 per year, AOC
is $10,000 per year, SL depreciation, i 12% per
year, T e 31%. Make the comparison using recovery
periods of 3 and 6 years.
17.25 Complete the last four columns of the table below
using an effective tax rate of 40% for an asset that
has a first cost of $20,000 and a 3-year recovery
period with no salvage value, using ( a ) straight
line depreciation and ( b ) MACRS depreciation.
All cash flows are in $1000 units.
Estimates, $
Year GI P OE D TI Taxes CFAT
0 — 20 — — — — 20
1 8 2
2 15 4
3 12 3
4 10 5
17.26 Use an effective tax rate of 32% to determine the
CFAT and NOPAT associated with the asset shown
below under two different scenarios: ( a ) with depreciation
at $6000 per year and ( b ) with depreciation
at $6000, $9600, $5760, and $3456 in years 1
through 4, respectively. All monetary amounts are
in $1000.
Estimates, $
Year GI OE P D TI Taxes CFAT NOPAT
0 — — 30 — — — 30 —
1 8 2
2 15 4
3 12 3
4 10 5
17.27 J. B. Hunt, Inc., an overland freight company, has
purchased new trailers for $150,000 and expects to
realize a net $80,000 in gross income over operating
expenses for each of the next 3 years. The trailers
have a recovery period of 3 years. Assume an
effective tax rate of 35% and an interest rate of
15% per year.
(a) Show the advantage of accelerated depreciation
by calculating the present worth of taxes
for the MACRS method versus the classical

476 Chapter 17 After-Tax Economic Analysis
SL method. Since MACRS takes an additional
year to fully depreciate the basis, assume
no CFBT beyond year 3, but include
any negative tax as a tax savings.
(b) Show that the total taxes are the same for
both methods.
17.28 A bioengineer is evaluating methods used to
apply adhesive onto microporous paper tape that
is commonly used after surgery. The machinery
costs $200,000, has no salvage value, and the
CFBT estimate is $75,000 per year for up to 10
years. The T e 38% and i 8% per year. The
two depreciation methods to consider are:
MACRS with n 5 years and SL with n
8 years (neglect the half-year convention effect).
For a study period of 8 years, ( a ) determine
which depreciation method offers the better tax
advantage, and ( b ) demonstrate that the same
total taxes are paid for MACRS and SL.
17.29 An asset with a first cost of $9000 is depreciated
by MACRS over a 5-year recovery period. The
CFBT is estimated at $10,000 for the first 4 years
and $5000 thereafter as long as the asset is retained.
The effective tax rate is 40%, and money is
worth 10% per year. In present worth dollars, how
much of the cash flow generated by the asset over
its recovery period is lost to taxes?
Depreciation Recapture and Capital Gains (Losses)
17.30 Last month, a company specializing in wind power
plant design and construction made a capital investment
of $400,000 in physical simulation
equipment that will be used for at least 5 years,
after which it is expected to be sold for approximately
25% of its first cost. According to tax law,
the simulation is MACRS-depreciated using a
3-year recovery period.
(a) Explain why there is a predictable tax implication
when the simulator is sold.
(b) Determine by how much the sale will cause
TI and taxes to change in year 5 if T e 35%.
The following information is used in Problems 17.31
through 17.34.
Open Access, Inc. is an international provider of computer
network communications gear. Different depreciation,
recovery period, and tax law practices in the three
countries where depreciable assets are located are summarized
in the table. Also, information is provided about
assets purchased 5 years ago at each location and sold this
year. After-tax MARR 9% per year and T e 30% can
be used for all countries.
Practice or Estimate Country 1 Country 2 Country 3
Depreciation
method
SL with
n 5
MACRS
with n 3
DDB with
n 5
Depreciation recapture Not taxed Taxed as TI Taxed as TI
First cost, $ 100,000 100,000 100,000
GI OE, $ per year 25,000 25,000 25,000
Estimated
salvage, $
0 in year 5 0 in year 5 20,000 in
year 5
Life, years 5 5 5
Actual selling
price, $
20,000 in
year 5
20,000 in
year 5
20,000 in
year 5
17.31 For country 1, SL depreciation is $20,000 per year.
Determine the ( a ) CFAT series and ( b ) PW of depreciation,
taxes, and CFAT series.
17.32 For country 2, MACRS depreciation for 4 years is
$33,333, $44,444, $14,815, and $7407, respectively.
Determine the ( a ) CFAT series and ( b ) PW
of depreciation, taxes, and CFAT series.
17.33 For country 3, DDB depreciation for 5 years is
$40,000, $24,000, $14,400, $1600, and 0, respectively.
Determine the ( a ) CFAT series and ( b ) PW
of depreciation, taxes, and CFAT series.
17.34 If you worked Problems 17.31 through 17.33, develop
a table that summarizes, for each country,
the total taxes paid and the PW values of the depreciation,
taxes, and CFAT series. For each criterion,
select the country that provides the best PW
value. Explain why the same country is not selected
for all three criteria. ( Hint : The PW should
be minimized for some criteria and maximized for
other criteria. Review previous material first to be
sure you choose correctly.)
17.35 An asset with a first cost of $350,000 three years
ago is sold for $385,000. The asset was depreciated
by the MACRS method and has a book value
of $100,800 at the time of sale. Determine the capital
gain and depreciation recapture, if any.
17.36 Sun-Tex Truck Stop is located in the desert southwest
and is 5 miles from the nearest municipal water
system. In order to have fresh water at the site, the
company purchased a turnkey reverse osmosis (RO)
system for $355,000. The company depreciated the
RO system using the MACRS method with a
10-year recovery period. Four years after the system
was purchased, water lines from a local water system
were extended to the truck stop, so Sun-Tex
sold the RO system for $190,000. Determine which
of the following apply and the amount, if any, to
include in an after-tax analysis of the project: depreciation
recapture, capital gain, capital loss. The
MACRS depreciation rates are 10%, 18%, 14.4%,
and 11.52% for years 1 through 4, respectively.

Problems 477
17.37 Freeman Engineering paid $28,500 for specialized
equipment for use with its new GPSGIS system.
The equipment was depreciated over a 3-year recovery
period using MACRS depreciation. The
company sold the equipment after 2 years for
$5000 when it purchased an upgraded system.
( a ) Determine the amount of the depreciation recapture
or capital loss involved in selling the asset.
( b ) What tax effect will this amount have?
17.38 Determine any depreciation recapture, capital gain,
or capital loss generated by each event described
below. Use them to determine the amount of income
tax effect, if the effective tax rate is 35%.
( a) A MACRS-depreciated asset with a 7-year
recovery period has been sold prematurely
after 4 years at an amount equal to 40% of its
first cost, which was $150,000.
( b) A hi-tech machine was sold internationally
for $10,000 more than its purchase price just
after it was in service 1 year. The asset had
P $100,000, S $1000, and n 3 years
and was depreciated by the MACRS method
for the 1 year.
( c) Land purchased 4 years ago for $1.8 million
was sold at a 10% profit.
( d) A 21-year-old asset was removed from
service and sold for $500. When purchased,
the asset was entered on the books with a
basis of P $180,000, S $5000, and n
18 years. Classical straight line depreciation
was used for the entire recovery period.
(e) A corporate car was depreciated using MACRS
over a 3-year recovery period. It was sold in
the fourth year of use for $2000.
17.39 Sunnen Products Co. of St. Louis, Missouri, makes
actuator hones for gas meter tubes where light-duty
metal removal is needed. The company purchased
land, a building, and two depreciable assets from
MPG Automation Systems Corporation, all of which
have recently been disposed of. Use the information
shown to determine the presence and amount of any
depreciation recapture, capital gain, or capital loss.
Asset
Purchase
Price, $
Recovery
Period,
Years
Current
Book
Value, $
Sales
Price, $
Land 220,000 — 295,000
Building 900,000 27.5 300,000 275,000
Asset 1 50,500 3 15,500 19,500
Asset 2 20,000 3 5,000 12,500
After-Tax Economic Analysis
17.40 Compute the required before-tax return if an aftertax
return of 7% per year is expected and the state
and local tax rates total 4.2%. The effective federal
tax rate is 34%.
17.41 Estimate the approximate after-tax rate of return
(ROR) for a project that has a before-tax ROR of
24%. Assume that the company has an effective
tax rate of 35% and it uses MACRS depreciation
for an asset that has a $27,000 salvage value.
17.42 Estimate the approximate after-tax rate of return
for a project that has a first cost of $500,000, a
salvage value of 20% of the first cost after 3 years,
and annual gross income less operating expenses
of (GIOE) $230,000. Assume the company
has a 35% effective tax rate.
17.43 An engineer is making an annual return of 12%
before taxes on the retirement investments placed
through her employer. No taxes are paid on retirement
earnings until they are withdrawn; however,
she was told by her brother, an accountant, that this
is the equivalent of an 8% per year after-tax return.
What percent of taxable income is her brother assuming
to be taken by income taxes?
17.44 Bart is an economic consultant to the textile industry.
In both a small business and a large corporation
he performed economic evaluations that have
an average 18% per year before-tax return. If the
stated MARR in both companies is 12% per year
after taxes, determine if management at both companies
should accept the projects. The before-tax
return is used to approximate the after-tax return.
Effective tax rates are 34% for the larger corporation
and 28% for the small company.
17.45 Elias wants to perform an after-tax evaluation of
equivalent methods A and B to electrostatically remove
airborne particulate matter from clean rooms
used to package liquid pharmaceutical products.
Using the information shown, MACRS depreciation
with n 3 years, a 5-year study period, after-tax
MARR 7% per year, and T e 34% and a spreadsheet,
he obtained the results AW A $−2176 and
AW B $3545. Any tax effects when the equipment
is salvaged were neglected. Thus, with MACRS depreciation,
method B is the better method. Now, use
classical SL depreciation with n 5 years to evaluate
the alternatives. Is the decision different from
that reached using MACRS?
Method A
Method B
First cost, $ 100,000 150,000
Salvage value, $ 10,000 20,000
Savings, $ per year 35,000 45,000
AOC, $ per year 15,000 6,000
Expected life, years 5 5

478 Chapter 17 After-Tax Economic Analysis
17.46 A corporation uses the following: before-tax
MARR of 14% per year, after-tax MARR of
7% per year, and T e of 50%. Two new machines
have the following estimates.
Machine A
Machine B
First cost, $ 15,000 22,000
Salvage value, $ 3,000 5,000
AOC, $ per year 3,000 1,500
Life, years 10 10
The machine is retained in use for a total of 10 years,
then sold for the estimated salvage value. Select one
machine under the following conditions:
(a) Before-tax PW analysis.
(b) After-tax PW analysis, using classical SL depreciation
over the 10-year life.
(c) After-tax PW analysis, using MACRS depreciation
with a 5-year recovery period.
17.47 Choose between alternatives A and B below if the
after-tax MARR is 8% per year, MACRS depreciation
is used, and T e 40%. The GIOE estimate
is made for only 3 years; it is zero when each asset
is sold in year 4.
Alternative
A
Alternative
B
First cost, $ 8,000 13,000
Actual salvage value, $ 0 2,000
Gross income expenses, $ 3,500 5,000
Recovery period, years 3 3
17.48 Offshore platform safety equipment, designed for
special jobs, will cost $2,500,000, will have no
salvage value, and will be kept in service for exactly
5 years, according to company policy. Operating
revenue minus expenses is estimated to be
$1,500,000 in year 1 and only $300,000 each additional
year. The effective tax rate for the multinational
oil company is 30%. Show hand and
spreadsheet solutions, as instructed, to find the
after-tax ROR using ( a ) classical SL depreciation
and ( b ) MACRS 5-year depreciation, neglecting
any tax effects at the end of 5 years of service.
17.49 Automatic inspection equipment purchased for
$78,000 by Stimson Engineering generated an average
of $26,080 annually in before-tax cash flow
during its 5-year estimated life. This represents a
return of 20%. However, the corporate tax expert
said the CFAT was only $15,000 per year. If the
corporation wants to realize an after-tax return of
10% per year, for how many more years must the
equipment remain in service?
After-Tax Replacement
17.50 In a replacement study between a defender and a
challenger, there may be a capital gain or loss
when the defender asset is sold. ( a ) How is the
gain or loss calculated, and ( b ) how does it affect
the AW values in the study?
17.51 In an after-tax replacement study of cost alternatives
involving one challenger and one defender,
how will a capital loss when selling the defender
affect the AW of each alternative?
17.52 An asset that was purchased 2 years ago was expected
to be kept in service for its projected life of 5 years, but
a new version (the challenger) of this asset promises
to be more efficient and have lower operating costs.
You have been asked to figure out if it would be more
economically attractive to replace the defender now
or keep it for 3 more years as originally planned. The
defender had a first cost of $300,000, but its market
value now is only $150,000. It has operating expenses
of $120,000 per year and no estimated salvage value
after 3 more years. To simplify calculations for this
problem only, assume that SL depreciation was
charged at $60,000 per year and that it will continue
at that rate for the next 3 years.
The challenger will cost $420,000, will have no
salvage value after its 3-year life, will have chargeable
expenses of $30,000 per year, and will be depreciated
at $140,000 per year (again, using SL
depreciation for simplicity in this case). Assume
the company’s effective tax rate is 35%, and its
after-tax MARR is 15% per year.
(a) Determine the CFAT in year 0 for the challenger
and defender.
(b)
Determine the CFAT in years 1 through 3 for
the challenger and defender.
(c) Conduct an AW evaluation to determine if
the defender should be kept for 3 more years
or replaced now.
17.53 The defender in a catalytic oxidizer manufacturing
plant has a market value of $130,000 and expected
annual operating costs of $70,000 with no salvage
value after its remaining life of 3 years. The depreciation
charges for the next 3 years will be $69,960,
$49,960, and $35,720. Using an effective tax rate
of 35%, determine the CFAT for next year only
that should be used in a present worth equation to
compare this defender against a challenger that
also has a 3-year life. Assume the company’s aftertax
MARR is 12%.
17.54 Perform a PW replacement study (hand and
spreadsheet solutions, if instructed) from the information
shown using an after-tax MARR 12%
per year, T e 35%, and a study period of 4 years.

Problems 479
(Assume that the assets will be salvaged at their
original salvage estimates. Since no revenues are
estimated, all taxes are negative and considered
“savings” to the alternative.) All monetary values
are in $1000 units.
Defender Challenger
First cost, $1000 45 24
Estimated S at purchase, $1000 5 0
Market value now, $1000 35 —
AOC, $1000 per year 7 8
Depreciation method SL MACRS
Recovery period, years 8 3
Useful life, years 8 5
Years owned 3 —
17.55 After 8 years of use, the heavy truck engine overhaul
equipment at Pete’s Truck Repair was evaluated
for replacement. Pete’s accountant used an
after-tax MARR of 8% per year, T e 30%, and a
current market value of $25,000 to determine
AW $2100. The new equipment costs $75,000,
uses SL depreciation over a 10-year recovery period,
and has a $15,000 salvage estimate. Estimated
CFBT is $15,000 per year. Pete asked his
engineer son Ramon to determine if the new equipment
should replace what is owned currently.
From the accountant, Ramon learned the current
equipment cost $20,000 when purchased and
reached a zero book value several years ago. Help
Ramon answer his father’s question.
17.56 Apple Crisp Foods signed a contract some years
ago for maintenance services on its fleet of
trucks and cars. The contract is up for renewal
now for a period of 1 year or 2 years only. The
contract quote is $300,000 per year if taken for
1 year and $240,000 per year if taken for 2 years.
The finance vice president wants to renew the
contract for 2 years without further analysis, but
the vice president for engineering believes it is
more economical to perform the maintenance
in-house. Since much of the fleet is aging and
must be replaced in the near future, a fixed
3-year study period has been agreed upon. The
estimates for the in-house (challenger) alternative
are as follows:
First cost, $ −800,000
AOC, $ per year −120,000
Life, years 4
Estimated salvage Loses 25% of P annually:
End year 1, S $600,000
End year 2, S $400,000
End year 3, S $200,000
End year 4, S $0
MACRS depreciation 3-year recovery period
The effective tax rate is 35%, and the after-tax
MARR is 10% per year. Perform an after-tax
AW analysis, and determine which vice president
has the better economic strategy over the
next 3 years.
17.57 Nuclear safety devices installed several years
ago have been depreciated from a first cost of
$200,000 to zero using MACRS. The devices
can be sold on the used equipment market for an
estimated $15,000. Or they can be retained in
service for 5 more years with a $9000 upgrade
now and an AOC of $6000 per year. The upgrade
investment will be depreciated over 3 years with
no salvage value. The challenger is a replacement
with newer technology at a first cost of
$40,000, n 5 years, and S 0. The new units
will have operating expenses of $7000 per year.
( a ) Use a 5-year study period, an effective tax
rate of 40%, an after-tax MARR of 12% per
year, and an assumption of classical straight line
depreciation (no half-year convention) to perform
an after-tax replacement study. ( b ) If the
challenger is known to be salable after 5 years
for an amount between $2000 and $4000, will
the challenger AW value become more or less
costly? Why?
17.58 Three years ago, Silver House Steel purchased a
new quenching system for $550,000. The
salvage value after 10 years at that time was estimated
to be $50,000. Currently the expected
remaining life is 7 years with an AOC of $27,000
per year. The new president has recommended
early replacement of the system with one that
costs $400,000 and has a 12-year life, a $35,000
salvage value, and an estimated AOC of $50,000
per year. The MARR for the corporation is 12%
per year. The president wishes to know the
replacement value that will make the recommendation
to replace now economically advantageous.
Use a spreadsheet and Solver to find the
minimum trade-in value ( a ) before taxes and
( b ) after taxes, using an effective tax rate of
30%. For solution purposes, use classical SL depreciation
for both systems. Comment on the
difference in replacement value made by the
consideration of taxes.
Economic Value Added
17.59 While an engineering manager may prefer to use
CFAT estimates to evaluate the AW of a project, a
financial manager may select AW of EVA estimates.
Why are these preferences predictable?
17.60 Cardenas and Moreno Engineering is evaluating
a very large flood control program for several

480 Chapter 17 After-Tax Economic Analysis
southern U.S. cities. One component is a 4-year
project for a special-purpose transport ship-crane
for use in building permanent storm surge protection
against hurricanes on the New Orleans coastline.
The estimates are P $300,000, S 0, and
n 3 years. MACRS depreciation with a 3-year
recovery is indicated. Gross income and operating
expenses are estimated at $200,000 and
$80,000, respectively, for each of 4 years. The
CFAT is shown below. Calculate the AW values
of the CFAT and EVA series. They should have
the same value. The after-tax MARR is 9.75%
and T e 35%.
Year GI, $ OE, $ P and S , $ D, $ TI, $ Taxes, $ CFAT, $
0 300,000 300,000
1 200,000 80,000 99,990 20,010 7,003 112,997
2 200,000 80,000 133,350 13,350 4,673 124,673
3 200,000 80,000 0 44,430 75,570 26,450 93,551
4 200,000 80,000 22,230 97,770 34,220 85,781
17.61 Triple Play Innovators Corporation (TPIC) plans
to offer IPTV (Internet Protocol TV) service to
North American customers starting soon. Perform
an AW analysis of the EVA series for the two alternative
suppliers available for the hardware and
software. Let T e 30% and after-tax MARR
8%; use SL depreciation (neglect half-year convention
and MACRS, for simplicity) and a study
period of 8 years.
Hong Kong
Vendor
Japan
First cost, $ 4.2 million 3.6 million
Recovery period, years 8 5
Salvage value, $ 0 0
GI − OE, $ per year 1,500,000 in year 1; increasing
by 300,000 per year up to 8 years
17.62 Sun Microsystems has developed partnerships
with several large manufacturing corporations to
use Java software in their consumer and industrial
products. A new corporation will be formed to
manage these applications. One major project involves
using Java in commercial and industrial
appliances that store and cook food. The gross
income and operating expenses are expected to
follow the relations shown for the estimated life of
6 years. For t 1 to 6 years,
Annual gross income, GI 2,800,000 100,000 t
Annual operating expenses, OE 950,000 50,000 t
The effective tax rate is 30%, the interest rate is
12% per year, and the depreciation method chosen
for the $3,000,000 capital investment is a 5-year
MACRS ADS alternative that allows straight line
write-off with the half-year convention in years 1
and 6. Using a spreadsheet, estimate ( a ) the annual
economic contribution of the project to the new
corporation and ( b ) the equivalent annual worth of
these contributions.
Value-Added Tax
17.63 What is the primary difference between a sales tax
and a value-added tax?
17.64 In Denmark, VAT is applied at a rate of 25%, with
few exceptions. Vendor A sells raw materials to
vendor B for $60,000 plus VAT, vendor B sells a
product to vendor C for $130,000 plus VAT, and
vendor C sells an improved, value-added product
to an end user for $250,000 plus VAT. Determine
the following.
(a) The amount of VAT collected by vendor B.
(b) The amount of tax vendor B sends to
Denmark’s Treasury.
(c) The total amount of tax collected by the
Treasury department.
The following information is used in Problems 17.65
through 17.70
Ajinkya Electronic Systems, a company in India that
manufactures many different electronic products, has to
purchase goods and services from a variety of suppliers
(wire, diodes, LED displays, plastic components, etc.).
The table below shows several suppliers and the VAT
rates associated with each. It also shows the purchases in
$1000 units that Ajinkya made (before taxes) from each
supplier in the previous accounting period. Ajinkya’s
sales to end users was $9.2 million, and Ajinkya’s products
carry a VAT of 12.5%.
Supplier VAT Rate, %
Purchases by
Ajinkya, $1000
A 4.0 350
B 12.5 870
C 12.5 620
D 21.3 90
E 32.6 50
17.65 How much VAT did supplier C collect?
17.66 How much tax did Ajinkya keep from the tax it
collected based on the purchases it made from
supplier A?

Additional Problems and FE Exam Review Questions 481
17.67 What was the total amount of taxes paid by Ajinkya
to the suppliers?
17.68 What was the average VAT rate paid by Ajinkya in
purchasing goods and services?
17.69 What was the amount of taxes Ajinkya sent to the
Treasury of India?
17.70 What was the total amount of taxes collected by
the Treasury of India from Ajinkya and Ajinkya’s
suppliers?
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
17.71 A graduated income tax system means:
( a) Only taxable incomes above a certain level
pay any taxes.
( b) A higher flat rate goes with all of the taxable
income.
( c) Higher tax rates go with higher taxable
incomes.
( d) Rates are indexed each year to keep up with
inflation.
17.72 All of the following are characteristics of a valueadded
tax system except:
( a) Value-added taxes are taxes on consumption.
( b) The end user pays value-added taxes.
( c) Value-added taxes are charged at each stage
of product development.
( d) Value-added taxes are charged only on the
raw materials for product development.
17.73 A small company has a taxable income that places
it in the 35% tax bracket. The amount of taxes that
a depreciation charge of $16,000 would save is
closest to:
( a ) $0
( b ) $3200
( c ) $5600
( d ) $10,400
17.74 A company that has a 50% effective tax rate had
income of $200 million in each of the last 2 years.
In one of those years, the company had deductions
of $100 million. In the other year, the company
had deductions of only $80 million. The difference
in income taxes paid by the company in those
2 years was closest to:
( a) $10 million
(b) $20 million
(c) $50 million
(d) $60 million
17.75 Taxable income (TI) is defined as:
( a) TI revenue operating expenses
depreciation
( b) TI revenue operating expenses
depreciation
(c)
(d)
TI revenue operating expenses
depreciation amortization
TI revenue operating expenses
depreciation
17.76 The marginal tax rate is defined as:
( a) The percentage paid on the last dollar of
i ncome
(b) The tax rate that applies to a questionable
investment
(c)
The tax rate that includes federal, state, and
local taxes
(d) The percentage paid on the first dollar of
income
17.77 A subcontractor with an effective tax rate of 25%
has gross income of $55,000, other income of
$4000, operating expenses of $13,000, and other
deductions and exemptions of $11,000. The income
tax due is closest to:
( a ) $11,750
( b ) $8,750
( c ) $10,750
( d ) $13,750
17.78 When a depreciable asset is disposed of for less than
its current book value, the transaction is known as:
( a) An after-tax expense
(b) Capital loss
(c) Capital gain
(d) Depreciation recapture
17.79 The after-tax analysis for a $60,000 investment
with associated gross income minus expenses
(GIOE) is shown below for the first 2 years only.
If the effective tax rate is 40%, the values for depreciation
( D ), taxable income (TI), and taxes for
year 1 are closest to:
Year
Investment,
$
GI − OE,
$ D, $ TI, $
Taxes,
$
CFAT,
$
0 60,000 60,000
1 30,000 26,000
2 35.000 15,000 6000 29,000

482 Chapter 17 After-Tax Economic Analysis
(a) D $5,000, TI $25,000, taxes $10,000
(b) D $30,000, TI $30,000, taxes $4,000
(c) D $20,000, TI $50,000, taxes $20,000
(d) D $20,000, TI $10,000, taxes $4,000
17.80 If the after-tax rate of return for a cash flow series
is 11.9% and the corporate effective tax rate is
34%, the approximated before-tax rate of return is
closest to:
( a ) 6.8%
( b ) 5.4%
( c ) 18.0%
( d ) 28.7%
17.81 An asset purchased for $100,000 with S $20,000
after 5 years was depreciated using the 5-year
MACRS rates. Expenses averaged $18,000 per
year, and the effective tax rate is 30%. The asset
was actually sold after 5 years of service for
$22,000. MACRS rates in years 5 and 6 are 11.53%
and 5.76%, respectively. The after-tax cash flow
from the sale is closest to:
( a ) $27,760
( b ) $17,130
( c ) $26,870
( d ) $20,585
17.82 When accelerated depreciation methods or
shortened recovery periods are applied, there
are impacts on the income taxes due. Of the following,
the statements that are commonly incorrect
are:
1. Total taxes paid are the same for all depreciation
methods.
2. Present worth of taxes is lower for shorter
recovery periods.
3. Accelerated depreciation imposes more taxes
in the later years of the recovery period.
4. Present worth of taxes is higher for shorter
recovery periods.
( a ) 1, 2, and 3
( b ) 1 and 4
( c ) 2
( d ) 4
CASE STUDY
AFTER-TAX ANALYSIS FOR BUSINESS EXPANSION
Background
Charles was always a hands-on type of person. Within a couple
of years of graduating from college, he started his own business.
After some 20 years, it has grown significantly. He owns and
operates Pro-Fence, Inc. in the Metroplex, specializing in custom-made
metal and stone fencing for commercial and residential
sites. For some time, Charles has thought he should expand
into a new geographic region, with the target area being another
large metropolitan area about 500 miles north, called Victoria.
Pro-Fence is privately owned by Charles; therefore, the question
of how to finance such an expansion has been, and still is,
the major challenge. Debt financing would not be a problem in
that the Victoria Bank has already offered a loan of up to $2 million.
Taking capital from the retained earnings of Pro-Fence is a
second possibility, but taking too much will jeopardize the current
business, especially if the expansion were not an economic
success and Pro-Fence were stuck with a large loan to repay.
This is where you come in as a long-time friend of
Charles. He knows you are quite economically oriented and
that you understand the rudiments of debt and equity financing
and economic analysis. He wants you to advise him on
the balance between using Pro-Fence funds and borrowed
funds. You have agreed to help him, as much as you can.
Information
Charles has collected some information that he shares with
you. Between his accountant and a small market survey of the
business opportunities in Victoria, the following generalized
estimates seem reasonable.
Initial capital investment $1.5 million
Annual gross income $700,000
Annual operating expenses $100,000
Effective income tax rate for Pro-Fence 35%
Five-year MACRS depreciation for all $1.5 million investment
The terms of the Victoria Bank loan would be 6% per year
simple interest based on the initial loan principal. Repayment
would be in 5 equal payments of interest and principal.
Charles comments that this is not the best loan arrangement
he hopes to get, but it is a good worst-case
scenario upon which to base the debt portion of the analysis.
A range of D-E mixes should be analyzed. Between
Charles and yourself, you have developed the following
viable options.
Debt
Loan
Percentage Amount, $
Equity
Investment
Percentage Amount, $
0 100 1,500,000
50 750,000 50 750,000
70 1,050,000 30 450,000
90 1,350,000 10 150,000

Case Study 483
Case Study Exercises
1. For each funding option, perform a spreadsheet analysis
that shows the total CFAT and its present worth over a
6-year period, the time it will take to realize the full
advantage of MACRS depreciation. An after-tax return
of 10% is expected. Which funding option is best for
Pro-Fence? ( Hint: For the spreadsheet, sample column
headings are: Year, GI − OE, Loan interest, Loan principal,
Equity investment, Depreciation rate, Depreciation,
Book value, TI, Taxes, and CFAT.)
2. Observe the changes in the total 6-year CFAT as the
D-E percentages change. If the time value of money
is neglected, what is the constant amount by which
this sum changes for every 10% increase in equity
funding?
3. Charles noticed that the CFAT total and PW values go in
opposite directions as the equity percentage increases.
He wants to know why this phenomenon occurs. How
should you explain this to Charles?
4. After deciding on the 50-50 split of debt and equity financing,
Charles wants to know what additional bottomline
contributions to the economic worth of the company
may be added by the new Victoria site. What are the best
estimates at this time?

CHAPTER 18
Sensitivity
Analysis and
Staged
Decisions
LEARNING OUTCOMES
Purpose : Perform a sensitivity analysis of parameters; use expected values to evaluate staged funding options.
SECTION TOPIC LEARNING OUTCOME
18.1 Sensitivity to variation • Use a measure of worth to explain sensitivity to
variation in one or more parameters.
18.2 Three-estimate variation • Choose the better alternative using three
estimates of variation in selected parameters.
18.3 E ( X ) • Calculate the expected value of a variable.
18.4 E ( X ) of cash flows • Evaluate a project or alternatives using expected
values of cash flows.
18.5 Decision trees • Use a decision tree to evaluate alternatives stage
by stage.
18.6 Real options • Explain a real option in engineering economics
and evaluate a staged decision using real
options analysis.

T
his chapter includes several related topics about alternative evaluation. Initially,
we expand our capability to perform a sensitivity analysis of one or
more parameters and of an entire alternative. Then the determination and use
of the expected value of a cash flow series are treated. The techniques of decision trees
help make a series of economic decisions for alternatives that have different, but closely
connected stages.
Economic decisions that involve staged funding are very common in professional and
everyday life. The last topic of real options analysis introduces a method useful in these
circumstances.
18.1 Determining Sensitivity to Parameter Variation
The term parameter is used in this chapter to represent any variable or factor for which an estimate
or stated value is necessary. Example parameters are first cost, salvage value, AOC, estimated
life, production rate, and materials costs. Estimates such as the loan interest rate and the
inflation rate can also be parameters of the analysis.
Economic analysis uses estimates of a parameter’s future value to assist decision makers.
Since future estimates are always incorrect to some degree, inaccuracy is present in the economic
projections. The effect of variation may be determined by using sensitivity analysis.
Sensitivity analysis determines how a measure of worth—PW, AW, FW, ROR, BC, or
CER—is altered when one or more parameters vary over a selected range of values. Usually
one parameter at a time is varied, and independence with other parameters is assumed. Though
this approach is an oversimplification in real-world situations, since the dependencies are difficult
to accurately model, the end results are usually correct.
Sensitivity analysis
In reality, we have applied this approach (informally) throughout previous chapters to determine
the response to variation in a variety of parameters. Variation in a parameter such as MARR will
not alter the decision to select an alternative when all compared alternatives return considerably
more than the MARR; thus, the decision is relatively insensitive to the MARR. However, variation
in the n or AOC value may indicate that the alternative’s measure of worth is very sensitive
to the estimated life or annual operating costs.
Usually the variations in life, annual costs, and revenues result from variations in selling price,
operation at different levels of capacity, inflation, etc. For example, if an operating level of 90%
of airline seating capacity for a domestic route is compared with 70% for a proposed international
route, the operating cost and revenue per passenger-mile will increase, but anticipated aircraft
life will probably decrease only slightly. Usually several important parameters are studied to
learn how the uncertainty of estimates affects the economic analysis.
Sensitivity analysis routinely concentrates on the variation expected in estimates of P , AOC,
S , n , unit costs, unit revenues, and similar parameters. These parameters are often the result of
design questions and their answers, as discussed in Chapter 15. Parameters that are interest rate–
based are not treated in the same manner.
Parameters such as MARR and other interest rates (loan rates, inflation rate) are more stable from
project to project. If performed, sensitivity analysis on them is for specific values or over a
narrow range of values. This point is important to remember if simulation is used for decision
making under risk (Chapter 19).
Plotting the sensitivity of PW, AW, or ROR versus the parameter(s) studied is very helpful.
Two alternatives can be compared with respect to a given parameter and the breakeven point.
This is the value at which the two alternatives are economically equivalent. However, the breakeven
chart commonly represents only one parameter per chart. Thus, several charts are constructed,
and independence of each parameter is assumed. In previous uses of breakeven analysis,
we often computed the measure of worth at only two values of a parameter and connected the
points with a straight line. However, if the results are sensitive to the parameter value, several
intermediate points should be used to better evaluate the sensitivity, especially if the relationships
are not linear.
When several parameters are studied, sensitivity analysis can become quite complex. It may
be performed one parameter at a time using a spreadsheet or computations by hand or calculator.
PW
Low
Parameter
High

486 Chapter 18 Sensitivity Analysis and Staged Decisions
EXAMPLE 18.1
The computer facilitates comparison of multiple parameters and multiple measures of worth, and
the software can rapidly plot the results.
Here is a general procedure to follow when conducting a thorough sensitivity analysis.
1. Determine which parameter(s) of interest might vary from the most likely estimated value.
2. Select the probable range and an increment of variation for each parameter.
3. Select the measure of worth.
4. Compute the results for each parameter, using the measure of worth as a basis.
5. To better interpret the sensitivity, graphically display the parameter versus the measure of
worth.
This sensitivity analysis procedure should indicate the parameters that warrant closer study
or require additional information. When there are two or more alternatives, it is better to use
the PW or AW measure of worth in step 3. If ROR is used, it requires the extra efforts of
incremental analysis between alternatives. Example 18.1 illustrates sensitivity analysis for
one project.
Wild Rice, Inc. expects to purchase a new asset for automated rice handling. Most likely estimates
are a first cost of $80,000, zero salvage value, and a cash flow before taxes (CFBT) per
year t that follows the relation $27,000 − 2000t. The MARR for the company varies over a
wide range from 10% to 25% per year for different types of investments. The economic life of
similar machinery varies from 8 to 12 years. Evaluate the sensitivity of PW by varying
(a) MARR, while assuming a constant n value of 10 years, and (b) n, while MARR is constant
at 15% per year. Perform the analysis by hand and by spreadsheet.
Solution by Hand
(a) Follow the procedure above to understand the sensitivity of PW to MARR variation.
1. MARR is the parameter of interest.
2. Select 5% increments to evaluate sensitivity to MARR; the range is 10% to 25%.
3. The measure of worth is PW.
4. Set up the PW relation for 10 years. When MARR 10%,
PW 80,000 25,000(PA,10%,10) 2000(PG,10%,10)
$27,830
The PW for all MARR values at 5% intervals is as follows:
MARR, % PW, $
10 27,830
15 11,512
20 962
25 10,711
5. A plot of MARR versus PW is shown in Figure 18–1. The steep negative slope indicates
that the decision to accept the proposal based on PW is quite sensitive to
variations in the MARR. If the MARR is established at the upper end of the range, the
investment is not attractive.
(b) 1. Asset life n is the parameter.
2. Select 2-year increments to evaluate PW sensitivity over the range 8 to 12 years.
3. The measure of worth is PW.
4. Set up the same PW relation as in part (a) at i 15%. The PW results are
n PW, $
8 7,221
10 11,511
12 13,145

18.1 Determining Sensitivity to Parameter Variation 487
30,000
20,000
MARR
Figure 18–1
Plot of PW versus MARR
and n for sensitivity
analysis, Example 18.1.
n
Present worth, $
10,000
0
10
6
15
8
20
10
25
12
MARR %
Life n
– 10,000
– 20,000
5. Figure 18–1 presents the plot of PW versus n. Since the PW measure is positive for all
values of n, the decision to invest is not materially affected by the estimated life. The PW
curve levels out above n 10. This insensitivity to changes in cash flow in the distant
future is a predictable observation, because the PF factor gets smaller as n increases.
Solution by Spreadsheet
Figure 18–2 presents two spreadsheets and accompanying plots of PW versus MARR (fixed n)
and PW versus n (fixed MARR). The NPV function calculates PW for i values from 10% to
25% and n values from 8 to 12 years. As the solution by hand indicated, so do the charts; PW
is sensitive to changes in MARR values, but not very sensitive to variations in n.
Figure 18–2
Sensitivity analysis of
PW to variation in
(a) MARR values and
(b) life estimates,
Example 18.1.
PW computation
NPV(C6,B$4:B$13)
B$3
(a)
PW computation
NPV(15%,B$4:B$15)
B$3
(b)

488 Chapter 18 Sensitivity Analysis and Staged Decisions
Figure 18–3
Sensitivity analysis graph
of percent variation from
the most likely estimate.
Rate of return on capital
50%
40%
30%
20%
10%
+
+
+
+
+
+
+
+
+
0%
– 10%
+
– 20%
– 50% – 40% – 30% – 20% – 10% 0% 10% 20% 30% 40% 50%
Parameter Legend
Sales price
Direct material
Sales volume
Indirect cost
Direct labor
Capital
% Change in individual parameter
When the sensitivity of several parameters is considered for one alternative using a single
measure of worth, it is helpful to graph percentage change for each parameter versus the measure
of worth. This is sometimes called a spider graph . Figure 18–3 illustrates ROR versus six
different parameters for one alternative. The variation in each parameter is indicated as a percentage
deviation from the most likely estimate on the horizontal axis. If the ROR response curve is
flat and approaches horizontal over the range of total variation graphed for a parameter, there is
little sensitivity of ROR to changes in the parameter’s value. This is the conclusion for indirect
cost in Figure 18–3 . On the other hand, ROR is very sensitive to sales price. A reduction of 30%
from the expected sales price reduces the ROR from approximately 20% to −10%, whereas a
10% increase in price raises the ROR to about 30%.
If two alternatives are compared and the sensitivity to one parameter is sought, the graph
may show quite nonlinear results. Observe the general shape of the sample sensitivity graphs in
Figure 18–4 . The plots are shown as linear segments between specific computation points. The
graph indicates that the PW of each plan is a nonlinear function of hours of operation. Plan A is
Figure 18–4
Sample PW sensitivity to
hours of operation for two
alternatives.
150
Plan B
PW, $ 1000
100
Plan A
50
0 1000 2000 3000
Hours of operation per year

18.1 Determining Sensitivity to Parameter Variation 489
very sensitive in the range of 0 to 2000 hours, but it is comparatively insensitive above
2000 hours. Plan B is more attractive due to its relative insensitivity. The breakeven point is at
about 1750 hours per year. It may be necessary to plot the measure of worth at intermediate
points to better understand the nature of the sensitivity.
EXAMPLE 18.2
Columbus, Ohio needs to resurface a 3-kilometer stretch of highway. Knobel Construction has
proposed two methods of resurfacing. The first method is a concrete surface for a cost of
$1.5 million and an annual maintenance cost of $10,000. The second method is an asphalt
covering with a first cost of $1 million and a yearly maintenance of $50,000. However, Knobel
requests that every third year the asphalt highway be touched up at a cost of $75,000.
The city uses the interest rate on bonds, 6% on its last bond issue, as the discount rate.
(a) Determine the breakeven number of years of the two methods. If the city expects an interstate
to replace this stretch of highway in 10 years, which method should be selected?
(b) If the touch-up cost increases by $5000 per kilometer every 3 years, is the decision sensitive
to this increase?
Solution
(a) Use PW analysis to determine the breakeven n value.
PW of concrete PW of asphalt
1,500,000 10,000(PA,6%,n) 1,000,000 50,000(PA,6%,n)
75,000 [ (PF,6%, j )
j
]
where j 3, 6, 9, . . . , n. The relation can be rewritten to reflect the incremental cash flows.
500,000 40,000(PA,6%,n) 75,000 [ (PF,6%, j )
j
] 0 [18.1]
The breakeven n value can be determined by hand solution by increasing n until Equation
[18.1] switches from negative to positive PW values. Alternatively, a spreadsheet solution
using the NPV function can find the breakeven n value (Figure 18–5). The NPV functions
in column C are the same each year, except that the cash flows are extended 1 year for each
Figure 18–5
Sensitivity of the breakeven
life between two alternatives,
Example 18.2.
PW for 11 years
NPV(6%,$B$5:$B15)$B$4
PW for 12 years
NPV(6%,$B$5:$B16)$B$4
Year counter advances by 1

490 Chapter 18 Sensitivity Analysis and Staged Decisions
present worth calculation. At approximately n 11.4 years, concrete and asphalt resurfacing
break even economically. Since the road is needed for 10 more years, the extra cost of
concrete is not justified; select the asphalt alternative.
(b) The total touch-up cost will increase by $15,000 every 3 years. Equation [18.1] is now
500,000 40,000(PA,6%,n) [ 75,000 15,000 ( ———
j 3
3 ) ] [
j
( PF,6%, j )
] 0
Now the breakeven n value is between 10 and 11 years—10.8 years using linear interpolation
(Figure 18–5, column E). The decision has become marginal for asphalt, since the
interstate is planned for 10 years hence.
Noneconomic considerations may be used to determine if asphalt is still the better alternative.
One conclusion is that the asphalt decision becomes more questionable as the
asphalt alternative maintenance costs increase; that is, the PW value is sensitive to increasing
touch-up costs.
EXAMPLE 18.3
18.2 Sensitivity Analysis Using Three Estimates
We can thoroughly examine the economic advantages and disadvantages among two or more
alternatives by borrowing from the field of project scheduling the concept of making three estimates
for each parameter: a pessimistic, a most likely, and an optimistic estimate. Depending
upon the nature of a parameter, the pessimistic estimate may be the lowest value (alternative life
is an example) or the largest value (such as asset first cost).
This approach allows us to study measure of worth and alternative selection sensitivity within a
predicted range of variation for each parameter. Usually the most likely estimate is used for all other
parameters when the measure of worth is calculated for one particular parameter or one alternative.
An engineer is evaluating three alternatives for new equipment at Emerson Electronics. She
has made three estimates for the salvage value, annual operating cost, and life. The estimates
are presented on an alternative-by-alternative basis in Table 18–1. For example, alternative B
has pessimistic estimates of S $500, AOC $4000, and n 2 years. The first costs are
known, so they have the same value. Perform a sensitivity analysis and determine the most
economical alternative, using AW analysis at a MARR of 12% per year.
TABLE 18–1
Competing Alternatives with Three Estimates Made for Salvage Value,
AOC, and Life Parameters
Strategy
First Cost,
$
Salvage Value S,
$
AOC,
$ per Year
Alternative A
P 20,000 0 11,000 3
Estimates ML 20,000 0 9,000 5
O 20,000 0 5,000 8
Alternative B
P 15,000 500 4,000 2
Estimates ML 15,000 1,000 3,500 4
O 15,000 2,000 2,000 7
Alternative C
P 30,000 3,000 8,000 3
Estimates ML 30,000 3,000 7,000 7
O 30,000 3,000 3,500 9
P pessimistic; ML most likely; O optimistic.
Life n,
Years

18.3 Estimate Variability and the Expected Value 491
TABLE 18–2 Annual Worth Values, Example 18.3
Alternative AW Values, $
Estimates A B C
P 19,327 12,640 19,601
ML 14,548 8,229 13,276
O 9,026 5,089 8,927
20
18
16
AW of costs, $1000
14
12
10
8
6
Alternative A
Alternative B
Alternative C
4
2
0
1
2
3
4
5
6
7
8
9
Life n, years
Figure 18–6
Plot of AW of costs for different-life estimates, Example 18.3.
Solution
For each alternative in Table 18–1, calculate the AW value of costs. For example, the AW relation
for alternative A, pessimistic estimates, is
AW 20,000(AP,12%,3) 11,000 $19,327
Table 18–2 presents all AW values. Figure 18–6 is a plot of AW versus the three estimates of
life for each alternative. Since the AW calculated using the ML estimates for alternative B
($−8229) is economically better than even the optimistic AW value for alternatives A and C,
alternative B is clearly favored.
Comment
While the alternative that should be selected here is quite obvious, this is not normally the case.
For example, in Table 18–2, if the pessimistic alternative B equivalent AW were much higher,
say, $21,000 per year (rather than $12,640), and the optimistic AW values for alternatives A
and C were less than that for B ($−5089), the choice of B would not be apparent or correct. In
this case, it would be necessary to select one set of estimates (P, ML, or O) upon which to base
the decision. Alternatively, the different estimates can be used in an expected value analysis,
which is introduced next.
18.3 Estimate Variability and the Expected Value
Engineers and economic analysts usually deal with estimate variation and risk about an uncertain
future by placing appropriate reliance on past data, if any exist. This means that probability and
samples are used. Actually the use of probabilistic analysis is not as common as might be expected.
The reason is not that the computations are difficult to perform or understand, but that
realistic probabilities associated with cash flow estimates are difficult to assign. Experience and

492 Chapter 18 Sensitivity Analysis and Staged Decisions
judgment can often be used in conjunction with probabilities and expected values to evaluate the
desirability of an alternative.
The expected value can be interpreted as a long-run average observable if the project is repeated
many times. Since a particular alternative is evaluated or implemented only once, the expected
value results in a point estimate. However, even for a single occurrence, the expected value is a
meaningful number.
The expected value E ( X ) is computed using the relation
im
E(X ) X i P(X i ) [18.2]
i1
where
X i value of the variable X for i from 1 to m different values
P ( X i ) probability that a specific value of X will occur
Probabilities are always correctly stated in decimal form, but they are routinely spoken of in
percentages and often referred to as chance , such as the chances are about 10%. When placing
the probability value in Equation [18.2] or any other relation, use the decimal equivalent of 10%,
that is, 0.1. In all probability statements the P ( X i ) values for a variable X must total to 1.0.
im
i1
P(X i ) 1.0
We may frequently omit the subscript i on X for simplicity.
If X represents the estimated cash flows, some will be positive and others will be negative. If
a cash flow sequence includes revenues and costs, and the measure of worth is present worth
calculated at the MARR, the result is the expected value of the discounted cash flows E (PW). If
the expected value is negative, the overall outcome is expected to be a cash outflow. For example,
if E (PW) $1500, this indicates that the proposal is not expected to return the MARR.
EXAMPLE 18.4
ANA airlines plans to offer several new electronic services on flights between Tokyo and selected
European destinations. The marketing director estimates that for a typical 24-hour period
there is a 50% chance of having a net cash flow of $5000 and a 35% chance of $10,000.
He also estimates there is a small 5% chance of no cash flow and a 10% chance of a loss of
$1000, which is the estimated extra personnel and utility costs to offer the services. Determine
the expected net cash flow.
Solution
Let NCF be the net cash flow in dollars, and let P (NCF) represent the associated probabilities.
Using Equation [18.2],
E (NCF) 5000(0.5) 10,000(0.35) 0(0.05) 1000(0.1) $5900
Although the “no cash flow” possibility does not increase or decrease E (NCF), it is included
because it makes the probability values sum to 1.0 and it makes the computation complete.
18.4 Expected Value Computations for Alternatives
The expected value computation E ( X ) is utilized in a variety of ways. Two prime ways are to:
• Prepare information for use in an economic analysis.
• Evaluate the expected viability of a fully formulated alternative.
Example 18.5 illustrates the first situation, and Example 18.6 determines the expected PW when
the entire cash flow series and its probabilities are estimated.

18.4 Expected Value Computations for Alternatives 493
EXAMPLE 18.5
There are many government incentives to become more energy-efficient. Installing solar
panels on homes, business buildings, and multiple-family dwellings is one of them. The
owner pays a portion of the total installation costs, and the government agency pays the
rest. Nichole works for the Department of Energy and is responsible for approving solar
panel incentive payouts. She has exceeded the annual budgeted amount of $50 million per
year in each of the previous 2 years. Disappointed with this situation, Nichole and her boss
decided to collect data to determine what size increase in annual budget the incentive program
needs in the future. Over the last 36 months, the amount of average monthly payout
and number of months are shown in Table 18–3 . She categorized by level the monthly
averages according to her experience with the program. Provided the same pattern continues,
what is the expected value of the dollar increase in annual budget that is needed to
meet the requests?
TABLE 18–3 Solar Panel Incentive Payouts, Example 18.5
Level
Average Payout,
$ Million per Month
Months over
Past 3 Years
Very high 6.5 15
High 4.7 10
Moderate 3.2 7
Low 2.9 4
Solution
Use the 36 months of payouts PO j ( j low, . . . ,very high) to estimate the probability P (PO j )
for each level, and make sure the total is 1.0.
Level, j Probability of Payout Level, P (PO j )
Very high
High
P (PO 1 ) 1536 0.417
P (PO 2 ) 1036 0.278
Moderate P (PO 3 ) 736 0.194
Low P (PO 4 ) 436 0.111
1.000
The expected monthly payout is calculated using Equation [18.2]. In $ million units,
E [PO] 6.5(0.417) 4.7(0.278) 3.2(0.194) 2.9(0.111)
2.711 1.307 0.621 0.322
$4.961 ($4,961,000)
The annual expected budget need is 12 4.961 million $59.532 million. The current budget
of $50 million should be increased by an average of $9.532 million per year.
EXAMPLE 18.6
Lite-Weight Wheelchair Company has a substantial investment in tubular steel bending equipment.
A new piece of equipment costs $5000 and has a life of 3 years. Estimated cash flows
( Table 18–4 ) depend on economic conditions classified as receding, stable, or expanding. A
probability is estimated that each of the economic conditions will prevail during the 3-year
period. Apply expected value and PW analysis to determine if the equipment should be purchased.
Use a MARR of 15% per year.

494 Chapter 18 Sensitivity Analysis and Staged Decisions
TABLE 18–4 Equipment Cash Flow and Probabilities, Example 18.6
Receding
(Prob. 0.4)
Solution
First determine the PW of the cash flows in Table 18–4 for each economic condition, and then
calculate E (PW) using Equation [18.2]. Define subscripts R for receding economy, S for stable,
and E for expanding. The PW values for the three scenarios are
PW R 5000 2500( PF ,15%,1) 2000( PF ,15%,2) 1000( PF ,15%,3)
5000 4344 $656
PW S 5000 5708 $708
PW E 5000 6309 $1309
Economic Condition
Stable
(Prob. 0.4)
Year Annual Cash Flow Estimates, $
Only in a receding economy will the cash flows not return the 15% to justify the investment.
The expected present worth is
E(PW) PW j [P(j)]
jR,S,E
656(0.4) 708(0.4) 1309(0.2)
$283
At 15%, E (PW) 0; the equipment is justified, using an expected value analysis.
Comment
It is also correct to calculate the E (cash flow) for each year and then determine PW of the
E (cash flow) series, because the PW computation is a linear function of cash flows. Computing
E (cash flow) first may be easier in that it reduces the number of PW computations. In this
example, calculate E (CF t ) for each year, then determine E (PW).
E (CF 0 ) $5000
E (CF 1 ) 2500(0.4) 2500(0.4) 2000(0.2) $2400
E (CF 2 ) $2400
E (CF 3 ) $2100
Expanding
(Prob. 0.2)
0 5000 5000 5000
1 2500 2500 2000
2 2000 2500 3000
3 1000 2500 3500
E (PW) 5000 2400( PF ,15%,1) 2400( PF ,15%,2) 2100( PF ,15%,3)
$283
18.5 Staged Evaluation of Alternatives
Using a Decision Tree
Alternative evaluation may require a series of decisions in which the outcome from one stage is important
to the next stage of decision making. When each alternative is clearly defined and probability estimates
can be made to account for risk , it is helpful to perform the evaluation using a decision tree .
A decision tree includes:
• More than one stage of alternative selection.
• Selection of an alternative at one stage that leads to another stage.
• Expected results from a decision at each stage.
• Probability estimates for each outcome.
• Estimates of economic value (cost or revenue) for each outcome.
• Measure of worth as the selection criterion, such as E(PW).

18.5 Staged Evaluation of Alternatives Using a Decision Tree 495
Decision
node
Alternatives
D
(a) Decision node
Probability
node
Outcomes
0.5
0.2
0.3
Probabilities
(b) Probability node with outcomes
D
D
D
Final outcomes
(c) Tree structure
Figure 18–7
Decision and probability nodes used to construct a decision tree.
The decision tree is constructed left to right and includes each possible decision and outcome.
• A square represents a decision node with the possible alternatives indicated on the branches
from the decision node ( Figure 18–7 a ).
• A circle represents a probability node with the possible outcomes and estimated probabilities
on the branches ( Figure 18–7 b ).
• The treelike structure in Figure 18–7 c results, with outcomes following a decision.
Usually each branch of a decision tree has some estimated economic value (often referred to
as payoff ) in cost, revenue, saving, or benefit. These cash flows are expressed in terms of PW,
AW, or FW values and are shown to the right of each final outcome branch. The cash flow and
probability estimates on each outcome branch are used in calculating the expected economic
value of each decision branch. This process, called solving the tree or rollback , is explained
after Example 18.7 , which illustrates the construction of a decision tree.
EXAMPLE 18.7
Jerry Hill is president and CEO of a U.S.-based food processing company, Hill Products and
Services. He was recently approached by an international supermarket chain that wants to
market in-country its own brand of frozen microwaveable dinners. The offer made to Jerry by
the supermarket corporation requires that a series of two decisions be made, now and 2 years
hence. The current decision involves two alternatives: (1) Lease a facility in the United Arab
Emirates (UAE) from the supermarket chain, which has agreed to convert a current processing
facility for immediate use by Jerry’s company; or (2) build and own a processing and packaging
facility in the UAE. Possible outcomes of this first decision stage are good market or poor
market depending upon the public’s response.
The decision choices 2 years hence are dependent upon the lease-or-own decision made
now. If Hill decides to lease, good market response means that the future decision alternatives
are to produce at twice, equal to, or one-half of the original volume. This will be a mutual decision
between the supermarket chain and Jerry’s company. A poor market response will indicate

496 Chapter 18 Sensitivity Analysis and Staged Decisions
a one-half level of production, or complete removal from the UAE market. Outcomes for the
future decisions are, again, good and poor market responses.
As agreed by the supermarket company, the current decision for Jerry to own the facility
will allow him to set the production level 2 years hence. If market response is good, the decision
alternatives are four or two times original levels. The reaction to poor market response
will be production at the same level or no production at all.
Construct the tree of decisions and outcomes for Hill Products and Services.
Solution
This is a two-stage decision tree that has alternatives now and 2 years hence. Identify the decision
nodes and branches, and then develop the tree using the branches and the outcomes of
good and poor market for each decision. Figure 18–8 details the decision stages and outcome
branches.
Stage 1 (decision now):
Label it D1.
Alternatives: lease (L) and own (O).
Outcomes: good and poor markets.
Good
Good
D2
2
1
0.5
Poor
Good
Poor
Good
Poor
Poor
D3
0.5
0
Good
Poor
Out of
business
Lease
(L)
D1
Own
(O)
Good
D4
4
2
Good
Poor
Good
Poor
Poor
Figure 18–8
A two-stage decision tree identifying alternatives and possible outcomes.
D5
1
0
Good
Poor
Out of
business

18.5 Staged Evaluation of Alternatives Using a Decision Tree 497
Stage 2 (Decisions 2 years hence):
Label them D2 through D5.
Outcomes: good market, poor market, and out of business.
Choice of production levels for D2 through D5:
Quadruple production (4); double production (2); level production
(1 ); one-half production (0.5); stop production (0)
The alternatives for future production levels (D2 through D5) are added to the tree and followed
by the market responses of good and poor. If the stop-production (0) decision is made
at D3 or D5, the only outcome is out of business.
To utilize the decision tree for alternative evaluation and selection, the following additional
information is necessary for each branch:
• The estimated probability that each outcome may occur. These probabilities must sum to 1.0
for each set of outcomes (branches) that result from a decision.
• Economic information for each decision alternative and possible outcome, such as initial
investment and estimated cash flows.
Decisions are made using the probability estimate and economic value estimate for each outcome
branch. Commonly the present worth at the MARR is used in an expected value computation
of the type in Equation [18.2]. This is the general procedure to solve the tree using PW
analysis:
1. Start at the top right of the tree. Determine the PW value for each outcome branch considering
the time value of money.
2. Calculate the expected value for each decision alternative.
E(decision) (outcome estimate)P(outcome) [18.3]
where the summation is taken over all possible outcomes for each decision alternative.
3. At each decision node, select the best E (decision) value—minimum cost or maximum value
(if both costs and revenues are estimated).
4. Continue moving to the left of the tree to the root decision in order to select the best alternative.
5. Trace the best decision path through the tree.
EXAMPLE 18.8
A decision is needed to either market or sell a new invention. If the product is marketed, the
next decision is to take it international or national. Assume the details of the outcome branches
result in the decision tree of Figure 18–9 . The probabilities for each outcome and PW of CFBT
(cash flow before taxes) are indicated. These payoffs are in millions of dollars. Determine the
best decision at the decision node D1.
Solution
Use the procedure above to determine that the D1 decision alternative to sell the invention
should maximize E (PW of CFBT).
1. Present worth of CFBT is supplied.
2. Calculate the expected PW for alternatives from nodes D2 and D3, using Equation [18.3].
In Figure 18–9 , to the right of decision node D2, the expected values of 14 and 0.2 in ovals
are determined as
E (international decision) 12(0.5) 16(0.5) 14
E (national decision) 4(0.4) 3(0.4) 1(0.2) 0.2
The expected PW values of 4.2 and 2 for D3 are calculated in a similar fashion.

498 Chapter 18 Sensitivity Analysis and Staged Decisions
Market
9
D1
High
6.16 0.2
0.8
Low
14
D2
4.2
D3
International
National
International
National
14
0.2
4.2
2
0.5
0.5
0.4
0.4
0.2
0.8
0.2
0.4
0.4
0.2
PW of
CFBT
($ million)
12
16
4
–3
–1
6
–3
6
–2
2
Sell
9
1.0
9
Expected values for each
decision alternative
Figure 18–9
Solution of a decision tree with present worth of estimated CFBT values,
Example 18.8 .
3. Select the larger expected value at each decision node. These are 14 (international) at D2
and 4.2 (international) at D3.
4. Calculate the expected PW for the two D1 branches.
E (market decision) 14(0.2) 4.2(0.8) 6.16
E (sell decision) 9(1.0) 9
The expected value for the sell decision is simple since the one outcome has a payoff of 9.
The sell decision yields the larger expected PW of 9.
5. The largest expected PW of CFBT path is to select the sell branch at D1 for a guaranteed
$9,000,000.
18.6 Real Options in Engineering Economics
As we learned with decision trees, many of the problems in engineering economy can be viewed
as staged decisions. When the decision to invest more or less can be delayed into the future, the
problem is called staged funding. As an illustration, assume a large company expects to sell an
energy-saving, window-mounted residential air conditioning unit at the rate of 100,000 per
month by the end of 2 years on the market. Decision makers may opt to (1) build the capacity to
supply100,000 per month to market immediately or (2) build capacity to supply 25,000 per
month now and test the market’s receptivity. If positive, they can stage the increase by 25,000
additional units each 6 months to meet current demand. Of course, if an aggressive competitor
enters the scene, or the economy falters, the staged funding decision will change as warranted.
These alternatives provide time-based options to the company. Before we go further, some definitions
are needed.
An option is a purchase or investment that contractually provides the privilege to take a
stated action by some stated time in the future, or the right to not accept the offer and forfeit
the option.

18.6 Real Options in Engineering Economics 499
A real option, in engineering economy terms, is the investment (cost) in a project, process, or
system. The options usually involve physical (real) assets, buildings, equipment, materials, and
the like, thus, the word real. Options may also be leases, subcontracts, or franchises. The investment
alternatives present varying amounts of risk, which is estimated by probabilities of occurrence
for predictable future events.
Real options analysis is the application of techniques to determine the economic consequences
of delaying the funding decisions as allowed by the option. The estimated cash flows and other
consequences of these delays are analyzed with risk taken into account to the degree possible. A
measure of worth, e.g., PW or AW, is the criterion used to make the staged funding decisions. A
decision may be to expand, continue as is, contract, abandon, or replicate the alternative at the
time the option must be exercised.
An inherent part of real options analysis is the uncertainty of future estimates, as it is for most
economic analyses. After some illustrations of real options, we will discuss the probabilistic
dimensions. Samples from industry and everyday personal life that can be formulated as real
o ptions follow.
Industrial Setting
New markets—Purchase equipment and staff to enter an expanding international market over
the next 5 years.
New planes—Purchase commercial airplanes now with an option to buy an additional 5 planes
over the next 3 years at the same price as that paid for the current order.
Removing car models—Ford Motor Company can decide to maintain production on an established
car model with dwindling sales for the next 3 years or can opt to discontinue the model
in stages over a 1- or 2-year period.
Drilling lease—Buy a drilling option contract from landowners to drill for oil and gas at some
time in the next 10 years. The drilling may not be justified at this time, but the contract offers
the option to drill were it to become economically advantageous based on events such as increased
oil prices or improved recovery technology.
Personal Decision Making
Extended car warranty—When purchasing a new car, the option to buy an extended-coverage
warranty beyond the manufacturer’s warranty is always an option. The price of the option is
the cost of the extended warranty. The uncertainties and risks are the future unknown costs for
repairs and failed components.
House insurance—When a homeowner has no mortgage to pay, maintaining house insurance
is an option. Deductibles are high enough, e.g., 1% to 5% of the fully appraised value, that
insurance primarily covers only catastrophic damage to the structure. Self-insurance, where
money is set aside for potential damages while accepting the risk that a major event will take
place, is an option for the homeowner.
Some of the primary characteristics (with an example) of a real options analysis performed within
the context of engineering economics are as follows:
• Cost to obtain the option to delay a decision (PW of initial investment, lease cost, or future
investment amount).
• Anticipated future options and cash flow estimates (double production with annual net cash
flows estimated).
• Time period for follow-on decisions (staged decision time, such as 1 year or a 3-year test period).
• Market and risk-free interest rates (expected market MARR of 12% per year and inflation
rate estimate of 4% per year).
• Estimates of risk and future uncertainty for each option (probability that an estimated cash
flow series will actually occur, if a specific option is selected).
• Economic criterion used to make a decision (PW, ROR, or other measure of worth).

500 Chapter 18 Sensitivity Analysis and Staged Decisions
It is common to use a decision tree to record and understand the options prior to performing a real
options analysis with risk included. Example 18.9 demonstrates the use of a decision tree and PW
analysis.
EXAMPLE 18.9
A start-up company in the solar energy production business, SolarScale Energy, Inc., has
developed and field-tested a modularized, scalar solar thermal electric (STE) generation
system that is relatively inexpensive to purchase and has an efficiency considerably better
than traditional photovoltaic (PV) panels. The technology is promising enough that Capital
Investor Funds (CIF) has provided $10 million for manufacturing. Additionally, a contract
with a consortium of sunbelt states has been offered, but not accepted thus far, for a total of
$1.5 million per year for a 2-year test period. By contract, the units will be marketed through
the state energy departments with all revenue going to the state treasuries. The lead engineer
at SolarScale, the manager of CIF, and a conservation representative for the state
consortium have developed the following staged-funding options, based on the delayed
decision to increase manufacturing production level until preliminary results of the 2-year
contract are in hand.
Condition Option CIF Funding Consortium Contract
Sales are excellent
( 5000 unitsyear)
2 production
level
Additional $10 million in
year 2
Additional 8 years;
$4 million in years 3–10
Sales are excellent
(3000–5000 unitsyear)
1 production
level
Nothing; no salvage after
10 years
Additional 8 years;
$1.5 million in
years 3–10
Additional 3 years;
$1.5 million in
years 3–5
Nothing
Sales are poor
(2000–3000 unitsyear)
Reduce to ½
production
Nothing; sell for
$2.5 million after 5 years
Sales are poor
( 2000 unitsyear)
Stop after
2 years
Nothing; sell for
$5 million after 2 years
(a) Develop the two-stage decision tree for the options described.
(b) The base case is the 1 production level with the 8-year follow-on contract from the consortium.
If the estimates for this option are considered the most likely (expected value)
estimates, determine the present worth at a MARR of 10% per year.
(c) Determine the PW values for each possible final outcome at 10% per year, and identify the
best economic option when the stage 2 funding decision must be made.
Solution
( a) Figure 18–10 details the options with the year shown at the bottom. There are 2 outcome
branches initially (accept option; decline option) and four final branches for the
accept decision at D1, based upon sales level. The decline option has a $0 outcome.
(SolarScale has other ways to pursue revenue that are not represented in this abbreviated
example.)
(b) Perform a PW evaluation as we have in all previous chapters, assuming the estimates are
point estimates over the 10-year life of the project. The resulting PW 1 0 as shown
below indicates the contract is not justified economically. In $ millions,
PW 1 10 1.5( PA ,10%,10)
$0.78 ($780,000)
(c) Figure 18–11 is a spreadsheet screen shot that calculates the PW for each option using the
NPV function. The i * values are also shown using the IRR function. Note that the sale of
production assets after 5 years for $2.5 million (½ level) or after 2 years for $5 million
(stop) is included. Also, the extra $10 million investment in year 2 for the 2 level option
is i ncluded.

18.6 Real Options in Engineering Economics 501
> 5000
2
Sales
excellent
D2
3000–5000
1
D1
Accept
option
Sales
poor
D3
1/2
2000–3000
< 2000
Stop
now
Decline
option
No
sales
Now
Year 2
Year 5 Year 10
Time
Figure 18–10
Decision tree showing real options over 10-year period, Example 18.9 .
Sale of assets in this year
Figure 18–11
PW analysis of real options without risk considered, Example 18.9 .
Only the 2 production-level option is justified at MARR 10% per year. If SolarScale
and CIF, the financial backers, are not convinced that the sales level will exceed
5000 units per year, the contract option should be declined. Some marketing survey information
and risk analysis may be very helpful before this important decision is made.
Of the characteristics listed above for real option situations, the primary one absent in
Example 18.9 is that of estimate variation and some measure of risk. Decision making under
risk is covered more extensively in the next chapter; however, we can use the following definition
for this discussion of real options analysis.
When a parameter can take on more than one value and there is any estimate of chance or
probability about the opportunity that each value may be observed, risk is present.
Risk

502 Chapter 18 Sensitivity Analysis and Staged Decisions
EXAMPLE 18.10
A coin has two sides. If it is a perfectly balanced coin, a flip of the coin should result in heads 50%
of the time and tails 50% of the time. If the coin is intentionally biased in weight, such that 58% of
the time it lands heads up, then the long-run probability of heads is P (heads) 0.58. Since the sum
of probabilities across all possible values must add to 1, the biased coin has P (tails) 0.42.
There are a couple of points worth mentioning about risk and the calculated PW values for real
options analysis.
• When the risk is higher and the stakes are larger, the real options analysis is often more valuable.
(We shall see this in Example 18.10 .)
• Since one of the objectives of real options analysis is to evaluate the economic consequences
of delaying a decision, a PW value of the base case that is moderately positive means that the
project is justified and should be accepted immediately, without the decision delay. On the
other hand, if the PW is largely negative , the delay is likely not worthwhile, as it would take
a very large positive PW to result in E (PW) 0. Thus, the project should be rejected now, not
delayed for a future decision.
We return to the previous example with some risk assessment added to determine if the contract
option should be declined, as indicated by the base case.
Before the stage 1 decision is made in Example 18.9 about SolarScale’s state-consortium contract
offer, some expected sales information was collected. Once the results were reviewed by
the three individuals in attendance at the final decision-making meeting—one representative
each for SolarScale, CIF, and the consortium—each person recorded her or his estimated probability
as a measure of risk that the sales would be excellent or poor, which represents the two
outcome possibilities were the option accepted. The results are as follows:
Probability of Outcome
Excellent
Poor
SolarScale 0.5 0.5
CIF 0.8 0.2
Consortium 0.6 0.4
Use these probability estimates to determine the expected PW value, provided equal weighting
is given to each representative’s input.
Solution
For each outcome (excellent and poor), select the best PW value from Figure 18–11 , and then
find E (PW) for each representative.
Excellent: From 2 level and 1 level, select 2 with PW $1.97 million.
Poor: From ½ level and stop now, select ½ with PW $2.76 million.
In $ million, E (PW) for each organization is
E (PW for SolarScale) 1.97(0.5) 2.76(0.5) $0.40
E (PW for CIF) 1.97(0.8) 2.76(0.2) $1.02
E (PW for consortium) 1.97(0.6) 2.76(0.4) $0.08
With a 1/3 chance assigned to each representative, the overall E (PW of stage 2 decision) is
E (PW of stage 2 decision) 0.33(0.40 1.02 0.08)
$0.23 ($230,000)
The base case of 1 level production in part ( b ) of Example 18.9 resulted in E (PW)
$−780,000. When compared with the positive E (PW) result here, we see that with consideration
of the different options of production level and probabilities for sales level, the expected
PW has increased to a positive value. All other things being equal, the state consortium offer
should be accepted; that is, accept the real option of the contract.

There are many other examples and dimensions of real options analysis in engineering economics
and in the area of financial analysis, where options analysis got its start some years ago.
If you are interested in this new and interesting area of analysis, consult more advanced texts and
journal articles on the topic of real options.
Problems 503
CHAPTER SUMMARY
In this chapter the emphasis is on sensitivity to variation in one or more parameters using a
specific measure of worth. When two alternatives are compared, compute and graph the
measure of worth for different values of the parameter to determine when each alternative is
better.
When several parameters are expected to vary over a predictable range, the measure of worth
is plotted and calculated using three estimates for a parameter—most likely, pessimistic, and
optimistic. This approach can help determine which alternative is best among several. Independence
between parameters is assumed in all these analyses.
The combination of parameter and probability estimates results in the expected value relation
E ( X ) XP ( X )
This expression is also used to calculate E (revenue), E (cost), E (cash flow), and E (PW) for the
entire cash flow sequence of an alternative.
Decision trees are used to make a series of alternative selections. This is a way to explicitly
take risk into account. It is necessary to make several types of estimates for a decision tree: outcomes
for each possible decision, cash flows, and probabilities. Expected value computations are
coupled with those for the measure of worth to solve the tree and find the best alternatives stage
by stage.
Staged funding over time can be approached using the evolving area of real options. Delaying
an investment decision and considering the risks of the future can improve the overall E (PW) of
a project, process, or system.
PROBLEMS
Sensitivity to Parameter Variation
18.1 Kahn Instruments is considering an investment of
$500,000 in a new product line. The company will
make the investment only if it will result in a rate
of return of 15% per year or higher. If the revenue
is expected to be between $135,000 and $165,000
per year for 5 years, determine if the decision to
invest is sensitive to the projected range of income
using a present worth analysis.
18.2 A young couple planning ahead for their retirement
has decided that $2,600,000 is the amount
they will need to retire comfortably 20 years
from now. For the past 5 years they have been
able to invest one of their salaries ($50,000 per
year, which includes employer contributions)
while living off the other one. They plan to start
a family sometime in the next 10 years, and
when they have their first child, one of the parents
will quit working, causing the savings to
decrease to $15,000 per year thereafter. If they
have gotten a rate of return of 10% per year on
their investments and expect to continue at this
ROR, is reaching their goal of $2.6 million
20 years from now sensitive to when they have
their first child (i.e., between now and 10 years
from now)? Use an FW analysis.
18.3 A company that manufactures high-speed submersible
rotary indexing spindles is considering
upgrading the production equipment to reduce
costs over a 6-year planning horizon. The company
can invest $80,000 now, 1 year from now,
or 2 years from now. Depending on when the investment
is made, the savings will vary. That is,
the savings will be $25,000, $26,000, or $29,000
per year if the investment is made now (year 0),
in 1 year, or in 2 years, respectively. Will the
timing of the investment affect the request to
make at least a 20% per year return? Use future
worth analysis.
18.4 Different membrane systems are under consideration
for treating 3 million gallons per day (MGD)
of cooling tower blowdown water to reduce its

504 Chapter 18 Sensitivity Analysis and Staged Decisions
volume. Option 1 is a low-pressure seawater reverse
osmosis (SWRO) system that will operate at
500 psi with a fixed cost of $465 per day and an
operating cost of $0.67 per 1000 gallons. A second
option is a higher-pressure SWRO system offered
by vendor X that operates at 800 psi and will have
a lower fixed cost of $328 per day (because of
fewer membranes); however, its operating cost
will be $1.35 per 1000 gallons. A third option is
also a high-pressure SWRO system from vendor Y,
who claims that its system will have a lower operating
cost of $1.28 per 1000 gallons and the same
fixed cost as that of vendor X. Determine if the
selection of a low- or high- pressure system is dependent
on the lower operating cost offered by
vendor Y.
18.5 A machine that is currently used in manufacturing
circuit board card locks has AW $–63,000 per
year. A possible replacement is under consideration
with a first cost of $64,000 and an operating
cost of $38,000 per year for the next 3 years. Three
different engineers have given their opinion, about
what the salvage value of the new machine will be
3 years from now: $10,000, $13,000, and $18,000.
Is the decision to replace the machine sensitive to
the salvage value estimates at the company’s
MARR of 15% per year?
18.6 An equipment alternative is being economically
evaluated separately by three engineers at
Raytheon. The first cost will be $77,000, and the
life is estimated at 6 years with a salvage value of
$10,000. The engineers disagree, however, on the
estimated revenue the equipment will generate.
Joe has made an estimate of $10,000 per year. Jane
states that this is too low and estimates $14,000,
while Carlos estimates $18,000 per year. If the
before-tax MARR is 8% per year, use PW to determine
if these different estimates will change the
decision to purchase the equipment.
18.7 The owner of a small construction company is
planning to purchase specialized equipment to
complete a contract he just received. The first
cost of the equipment is $250,000, and it will
likely have a salvage value of $90,000 in 3 years,
at which time he will not need the equipment
anymore. The operating cost is expected to be
$75,000 per year. Alternatively, the owner can
subcontract the work for $175,000 per year. Because
the equipment is specialized, the owner is
not sure about the salvage value. He thinks it
might be worth as little as $10,000 in 3 years
(a scrap value). If his minimum attractive rate of
return is 15% per year, determine if the decision
to buy the equipment is sensitive to the salvage
value.
18.8 A company planning to borrow $10.5 million for a
plant expansion is not sure what the interest rate
will be when it applies for the loan. The rate could
be as low as 10% per year or as high as 12% per
year for a 5-year loan. The company will only
move forward with the project if the annual worth
of the expansion is below $5.7 million. The M&O
cost is fixed at $3.1 million per year. The salvage
could be $2 million if the interest rate is 10% or
$2.5 million if it is 12% per year. Is the decision to
move forward with the project sensitive to the interest
rate and salvage value estimates?
18.9 A company that manufactures clear PVC pipe is
investigating the production options of batch and
continuous processes. Estimated cash flows are as
follows:
Batch
Continuous
First cost, $ 80,000 130,000
Annual cost, $ per year 55,000 30,000
Salvage value for any year, $ 10,000 40,000
Life, years 3–10 5
The chief operating officer (COO) has asked you
to determine if the batch option would ever have a
lower annual worth than the continuous flow system,
using interest rates over a range of 5% to 15%
for the batch option but only 15% for the continuous
flow system. ( Note : The continuous flow process
was previously determined to have its lowest
cost over a 5-year life cycle; the batch process can
be used from 3 to 10 years.)
18.10 An engineer collected average cost and revenue data
for Arenson’s FC1 handheld financial calculator.
Fixed cost $300,000 per year
Cost per unit $40
Revenue per unit $70
( a) What is the range in breakeven quantity if
there is possible variation in the fixed cost
from $200,000 to $400,000 per year? (Use
$50,000 increments.)
( b) What is the incremental change in the breakeven
quantity for each $50,000 change in
fixed cost?
The following information is used for Problems 18.11
through 18.14.
A new online patient diagnostics system for surgeons will
cost $200,000 to install, cost $5000 annually to maintain
and will have an expected life of 5 years. The added revenue
is estimated to be $60,000 per year, and the MARR
is 10% per year. Examine the sensitivity of present worth
to variation in selected parameter estimates, while others
remain constant.

Problems 505
18.11 Sensitivity to first cost variation: $150,000 to
$250,000 (−25% to 25%).
18.12 Sensitivity to revenue variation: $45,000 to $75,000
(−25% to 25%).
18.13 Sensitivity to life variation: 4 years to 7 years (−20%
to 40%).
18.14 Plot the results on a graph similar to Figure 18–3
and comment on the relative sensitivity of each
parameter.
18.15 Charlene plans to place an annual savings
amount of A $27,185 into a retirement program
at the end of each year for 20 years starting
next year. She expects to retire and start to draw
a total of R $60,000 per year 1 year after the
20th deposit. Assume an effective earning rate
of i 6% per year on the retirement investments
and an infinite life. Determine and comment on
the sensitivity of the size of the annual withdrawal
R for variations in A and i. Show hand
and spreadsheet solutions.
( a) Variation of 5% in the annual deposit A.
(b) Variation of 1% in the effective earning rate
i , that is, ranging from 5% to 7% per year.
18.16 Ned Thompson Labs performs tests on super
alloys, titanium, aluminum, and most metals. Tests
on metal composites that rely upon scanning electron
microscope results can be subcontracted, or
the labs can purchase new equipment. Evaluate the
sensitivity of the economic decision to purchase
the equipment over a range of 20% (in 10% increments)
of the estimates for P , AOC, R , n , and
MARR (range on MARR is 12% to 16%). Use the
AW method and plot the results on a sensitivity
graph (like Figure 18–3 ). For which parameter(s)
is the AW most sensitive? Least sensitive?
First cost P $−220,000
Salvage S $20,000
Life n 10 years
Annual operating cost AOC $−30,000/year
Annual revenue R $70,000 per year
MARR i 15% per year
18.17 Titan manufactures and sells gas-powered electricity
generators. It can purchase a new line of fuel
injectors from either of two companies. Cost and
savings estimates are made, but the savings estimate
is unreliable at this time. Use an AW analysis
at 10% per year to determine if the selection between
company A and company B changes when
the savings per year may vary as much as 40%
from the best estimates made thus far.
Company A
Company B
First cost, $ 50,000 37,500
AOC, $ per year 7,500 8,000
Savings best estimate, $ per year 15,000 13,000
Salvage, $ 5,000 3,700
Life, years 5 5
18.18 ( a) Graph the sensitivity of what a person
should be willing to pay now for a 9%,
$10,000 bond due in 10 years if there is a
30% change in (1) face value, (2) dividend
rate, or (3) required nominal rate of return,
which is expected to be 8% per year, compounded
semiannually. The bond pays dividends
semiannually.
( b) If the investor did purchase the $10,000 face
value bond at a premium of 5% (i.e., 5%
above face value) and all your other estimates
were correct, that is, 0% change, did
he pay too much or too little? How much?
Three Estimates
18.19 DVH Technologies purchases several parts for
the instruments it makes via a fixed-price contract
of $190,000 per year from a local supplier.
The company is considering making the parts
in-house through the purchase of equipment that
will have a first cost of $240,000 with an estimated
salvage value of $30,000 after 5 years. The
operating cost is difficult to estimate, but company
engineers have made optimistic, most likely,
and pessimistic estimates of $60,000, $85,000,
and $120,000 per year, respectively. Determine if
the company should purchase the equipment
under any of the operating cost scenarios. The
MARR is 20% per year.
18.20 Astor Engineering recently merged with another
firm and could lease additional office space or
purchase its own building. The $30,000 per year
lease agreement will be a net, net, net lease,
which means that the lessee (Astor) will pay the
real estate taxes on the leased space, the building
insurance on the leased space, and the common
area maintenance. Since these costs are about the
same if Astor owned the building, they do not
need to be considered in the analysis. A new
building will cost $880,000 to purchase, but there
is considerable uncertainty about what it will be
worth in 20 years, which is the planning period
selected. The individuals involved in the discussion
made optimistic, most likely, and pessimistic
estimates of $2,400,000, $1,400,000, and
$900,000, respectively. Determine if Astor should
purchase the building under any of the estimated
resale values at i 10% per year.

506 Chapter 18 Sensitivity Analysis and Staged Decisions
18.21 Holly Farms is considering two environmental
chambers to accomplish detailed laboratory confirmations
of online bacteria tests in chicken
meat for the presence of E. coli 0157:H7 and Listeria
monocytogenes . There is some uncertainty
about how long the D103 chamber will be useful.
A realistic estimate is 3 years, but pessimistic
and optimistic estimates of 2 years and 6 years,
respectively, are also possible. The estimated
salvage value will remain the same. Using an interest
rate of 10% per year, determine if any of
the D103 estimates would result in a lower cost
than that of the 490G chamber for a 6-year planning
period.
Chamber D103 Chamber 490G
Installed cost, $ 400,000 250,000
AOC, $ per year 4,000 3,000
Salvage value at 10% of P , $ 40,000 25,000
Life, years 2, 3, or 6 2
18.22 When the country’s economy is expanding, AB Investment
Company is optimistic and expects a
MARR of 15% for new investments. However, in
a receding economy the expected return is 8%.
Normally a 10% return is required. An expanding
economy causes the estimates of asset life to go
down about 20%, and a receding economy makes
the n values increase about 10%. Calculate and observe
or plot the sensitivity of PW values versus
( a ) the MARR and ( b ) the life values for the two
plans detailed below, using the most likely estimates
for the other factors. ( c ) Considering all the
analyses, under which scenario, if any, should plan
M or Q be rejected?
Plan M
Plan Q
Initial investment, $ 100,000 110,000
Cash flow, $ per year 15,000 19,000
Life, years 20 20
Expected Value
18.23 Determine the expected net operating income
(NOI) from sales of micro turbine components.
The probabilities are 20%, 50%, and 30% for revenues
of $800,000, $1,000,000, and $1,100,000
per year, respectively, and operating expenses are
constant at $200,000 per year.
18.24 A company that manufactures amplified pressure
transducers is trying to decide between a dualspeed
and a variable-speed machine. The engineers
are not sure about the salvage value of the
variable-speed machine, so they have asked several
different used-equipment dealers for
estimates. The results can be summarized as follows:
there is a 32% chance of getting $20,000, a
45% chance of getting $28,000, and a 13%
chance of getting $34,000. Also, there is a 10%
chance that the company may have to pay $5000
to dispose of the equipment. Calculate the
expected salvage value.
18.25 The average success probability for a wildcat oil
well drilled in the Wind River basin 7 miles from
the nearest existing production well is estimated
to be 13%. If the value of the oil has equal
chances of being $1.5 million, $1.9 million, and
$2.4 million, what is the expected income from
the well?
18.26 Nationwide income from monthly sales data
(rounded to the nearest $100,000) of Stay Flat
vacuum hold-down tables for last year is shown
below. Determine the expected value of the
monthly income, if economic conditions remain
the same.
Income, $ per Month
Number
of Months
500,000 4
600,000 2
700,000 1
800,000 2
900,000 3
18.27 Determine the expected maximum rainfall intensity
in El Paso, Texas for the month of July using
the estimated probabilities shown.
Rainfall Rate, inches per hour 3 4 5 6
Probability 0.4 0.3 0.2 0.1
18.28 There are four estimates made for the anticipated
cycle time to produce a subcomponent. The estimates,
in seconds, are 10, 20, 30, and 50. ( a ) If
equal weight is placed on each estimate, what is
the expected cycle time? ( b ) If the largest time is
disregarded, what is the percent reduction in the
expected time?
18.29 The PW value for an alternative is expected to be
one of two values based on bids from two vendors.
Your office partner told you that the low bid is
$3200 per year. If she indicates a chance of 70% of
accepting the high bid and that her expected PW is
$5875, what is the PW of the high bid?
18.30 A total of 40 different proposals were evaluated by
the IRAD (Industrial Research and Development)
committee during the past year. Twenty were
funded. Their rate of return estimates are summarized
with the i * values rounded to the nearest

Problems 507
integer. For the accepted proposals, calculate the
expected rate of return E ( i ).
Proposal
ROR, i *%
Number of
Proposals
8 1
5 1
0 5
5 5
8 2
10 3
15 3
20
18.31 Beckman Electronics has performed an economic
analysis of proposed service in a new region of the
country. The three-estimate approach to sensitivity
analysis has been applied. The optimistic and pessimistic
values each have an estimated 20% chance
of occurring. Use the FW values shown to determine
the expected FW.
Optimistic Most Likely Pessimistic
FW value, $ 300,000 50,000 25,000
18.32 A very successful health and recreation club
wants to construct a mock mountain for climbing
and exercise outside for its customers’ use.
Because of its location, there is a 30% chance of
a 120-day season of good outdoor weather, a
50% chance of a 150-day season, and a 20%
chance of a 165-day season. The mountain will
be used by an estimated 350 persons each day of
the 4-month (120-day) season, but by only 100
per day for each extra day the season lasts. The
feature will cost $375,000 to construct and require
a $25,000 rework each 4 years; and the annual
maintenance and insurance costs will be
$56,000. The climbing fee will be $5 per person.
If a life of 10 years is anticipated and a 12% per
year return is expected, determine if the addition
is economically justified.
18.33 The owner of Ace Roofing may invest $200,000 in
new equipment. A life of 6 years and a salvage
value of 12% of first cost are anticipated. The annual
extra revenue will depend upon the state of
the housing and construction industry. The extra
revenue is expected to be only $20,000 per year if
the current slump in the industry continues. Real
estate economists estimate a 50% chance of the
slump lasting 3 years and they give it a 20% chance
of continuing for 3 additional years. However, if
the depressed market does improve, during either
the first or second 3-year period, the revenue of the
investment is expected to increase by a total of
$35,000 per year. Can the company expect to make
a return of 8% per year on its investment? Use
present worth analysis.
18.34 A flagship hotel in Cedar Falls must construct a
retaining wall next to its parking lot due to the
widening of the city’s main thoroughfare located
in front of the hotel. The amount of rainfall experienced
in a short time may cause damage in varying
amounts, and the wall increases in cost in
order to protect against larger and faster rainfalls.
The probabilities of a specific amount of rainfall
in a 30-minute period and wall cost estimates are
as follows:
Rainfall, Inches
per 30 Minutes
Probability of
Greater Rainfall
First Cost
of Wall, $
2.0 0.3 200,000
2.25 0.1 225,000
2.5 0.05 300,000
3.0 0.01 400,000
3.25 0.005 450,000
The wall will be financed through a 6% per year
loan. The principal and interest will be repaid over
a 10-year period. Records indicate an average
damage of $50,000 has occurred with heavy rains,
due to the relatively poor cohesive properties of
the soil along the thoroughfare. A discount rate of
6% per year is applicable. Find the amount of rainfall
to protect against by choosing the retaining
wall with the smallest AW value over the 10-year
period.
Decision Trees
18.35 For the decision tree branch shown, determine the
expected values of the two outcomes if decision
D3 is already selected and the maximum outcome
value is sought. (This decision branch is part of a
larger tree.)
D3
Probability
0.4
0.3
0.3
0.6
0.4
Value, $
55
–30
10
–17
18.36 A large decision tree has an outcome branch detailed
(next page). If decisions D1, D2, and D3 are
all options in a 1-year period, find the decision
0

508 Chapter 18 Sensitivity Analysis and Staged Decisions
path that maximizes the outcome value. There are
specific investments necessary for decision nodes
D1, D2, and D3, as indicated on each branch.
Investment
D1
$50
$25
0.9
0.5
D2 $30 0.5
$20
$80
0.1
0.3
0.3
0.4
D3
Value $
150
–30
75
200
– 100
50
0.4
Value, $
30
100
0.6 –50
500
18.37 Decision D4, which has three possible alternatives—
x , y , or z —must be made in year 3 of a 6-year
study period in order to maximize the expected
value of present worth. Using a rate of return of 15%
per year, the investment required in year 3 and the
estimated cash flows for years 4 through 6, determine
which decision should be made in year 3.
D4
x
y
z
High
Low
High
Low
High
Low
Investment Cash flow, $1000
Required, $
Years
3 4 5 6
– 200,000
– 75,000
50
40
30
30
– 350,000 190
–30
50
30
40
30
170
–30
50
20
50
30
150
–30
90
Outcome
probability
0.7
0.3
0.45
0.55
0.7
0.3
18.38 A total of 5000 mechanical subassemblies are
needed annually on a final assembly line. The subassemblies
can be obtained in one of three ways:
(1) Make them in one of three plants owned by the
company; (2) buy them off the shelf from the one
and only manufacturer; or (3) contract to have them
made to specifications by a vendor. The estimated
annual equivalent cost for each alternative is dependent
upon specific circumstances of the plant, producer,
or contractor. The information shown details
the circumstance, a probability of occurrence, and
the estimated annual cost. Construct and solve a decision
tree to determine the least-cost alternative to
provide the subassemblies.
Decision
Alternative Outcomes Probability
Annual Cost for
5000 Units, $
per Year
1. Make Plant:
A 0.3 250,000
B 0.5 400,000
C 0.2 350,000
2. Buy off
the shelf
Quantity:
5000, pay premium 0.2 550,000
5000 available 0.7 250,000
5000, forced to buy 0.1 290,000
3. Contract Delivery:
Timely delivery 0.5 175,000
Late delivery, then
buy some off shelf
0.5 450,000
18.39 The president of ChemTech is trying to decide
whether to start a new product line or purchase a
small company. It is not financially possible to do
both. To make the product for a 3-year period will
require an initial investment of $250,000. The expected
annual cash flows with probabilities in
parentheses are: $75,000 (0.5), $90,000 (0.4), and
$150,000 (0.1). To purchase the small company
will cost $450,000 now. Market surveys indicate a
55% chance of increased sales for the company and
a 45% chance of severe decreases with an annual
cash flow of $25,000. If decreases are experienced
in the first year, the company will be sold immediately
(during year 1) at a price of $200,000. Increased
sales could be $100,000 the first 2 years. If
this occurs, a decision to expand after 2 years at an
additional investment of $100,000 will be considered.
This expansion could generate cash flows
with indicated probabilities as follows: $120,000
(0.3), $140,000 (0.3), and $175,000 (0.4). If expansion
is not chosen, the current size will be maintained
with anticipated sales to continue. Assume
there are no salvage values on any investments. Use
the description given and a 15% per year return to
do the following.
(a) Construct a decision tree with all values and
probabilities shown.
(b) Determine the expected PW values at the
“ expansion/no expansion” decision node after
2 years, provided sales are up.
(c) Determine what decision should be made
now to offer the greatest return possible for
ChemTech.

Additional Problems and FE Exam Review Questions 509
( d) Explain in words what would happen to the
expected values at each decision node if the
planning horizon were extended beyond
3 years and all cash flow values continued as
forecasted in the description.
Real Options
18.40 A privately held company that makes chips that are
essential for high-volume data storage is valued at
$3 billion. A computer company that wants to get
into cloud computing is considering purchasing
the company, but because of the uncertain economy,
it would prefer to purchase an option that will
allow it to buy the company for up to 1 year from
now at a cost of $3.1 billion. What is the maximum
amount the company should be willing to pay for
the option, if its MARR is 12% per year?
18.41 A company that is considering adding a new product
line has determined that the first cost would be
$80 million. The company is not sure about how
the product will be received, so it has projected revenues
using optimistic, most likely, and pessimistic
estimates of $35 million, $25 million, and $10 million,
respectively, with equal probability for each.
Instead of expanding now, the company could implement
a test program for 1 year in a limited area
that will cost $4 million. (The full-scale project will
still cost $80 million if implemented after the test
program is over.) This will provide the company
with the option to move forward or cancel the project.
The criterion identified to move ahead with
full-scale implementation is that revenues must exceed
$900,000. In this case, the pessimistic estimate
will be eliminated, and equal probability will
be placed on the remaining revenue projections. If
the company uses a 5-year planning horizon and a
MARR of 12% per year, should the company go
ahead with the full-scale project now or take the
option to implement the test program for 1 year?
18.42 Dow Chemical is considering licensing a low liquid
discharge (LLD) water treatment system from a
small company that developed the process. Dow
can purchase a 1-year option for $150,000 that will
give it time to pilot-test the LLD process, or Dow
can acquire the license now at a cost of $1.8 million
plus 25% of sales. If Dow waits 1 year, the cost
will increase to $1.9 million plus 30% of sales. If
Dow projects the sales to be $1,000,000 per year
over the 5-year license period, should the company
license the process now or purchase the option to
license it after the 1-year test period? Assume the
MARR is 15% per year.
18.43 Abby has just negotiated a $15,000 price on a
2-year-old car and is with the salesman closing the
deal. There is a 1-year sales warranty with the purchase;
however, an extended warranty is available
for $2500 that will cover the same repairs and component
failures as the 1-year warranty for 3 additional
years. Abby understands this to be a real options
situation with the price of the option ($2500)
paid to avoid future, unknown costs. To help with
her decision, the salesman provided three typical
sets of historical data on estimated repair costs for
used cars. The first-year costs are shown as zero
because they will be covered by the sales warranty.
Year
1 2 3 4
Repair cost, $ per year:
A 0 500 1200 850
B 0 1000 1400 400
C 0 0 500 2000
The salesman said case C is the base case, since it
shows that the extended warranty is not needed because
the cost of repairs equals the warranty cost.
Abby immediately recognized this to be the case
only when i 0%.
(a)
If Abby assumes that each repair cost scenario
has equal probability of occurring with her
car, and money is worth 5% per year to her,
how much should she be willing to pay for the
extended warranty that is offered at $2500?
(b) If the base case actually occurs for her car
and she does not purchase the warranty, what
is the PW value of the expected future costs
at i 5% per year?
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
18.44 In conducting a sensitivity analysis, all of the
following could be used as a measure of worth
except:
( a) Present worth
( b) Cost-capacity equations
( c) Annual worth
( d) Benefit-cost ratio
18.45 When the measure of worth is plotted versus percent
change for several parameters, the parameter that is
the most sensitive in the economic analysis is the one:
(a) That has the steepest curve
(b) That has the flattest curve
(c) With the largest present worth
(d) With the shortest life

510 Chapter 18 Sensitivity Analysis and Staged Decisions
18.46 In conducting a formalized sensitivity analysis
using three estimates, the three estimates should be:
(a) Strategic, pessimistic, real likely:
(b) Deterministic, realistic, optimistic
(c) Optimistic, pessimistic, most likely
(d) Authentic, realistic, probabilistic
18.47 For annual worth values of $30,000, $40,000, and
$50,000 with chances of 20%, 20%, and 60%, respectively,
the expected AW is closest to:
(a) $34,000
(b) $40,000
(c) $44,000
(d) $48,000
18.48 The AW of a 3-year-old machine that is used in the
manufacture of modular magnetic encoders, which
provide data on the speed and positioning of rotating
motor shafts, is $−48,000. A challenger will have a
first cost of $90,000, an operating cost of $29,000,
and a salvage value after 5 years that may vary considerably.
For an interest rate of 10% per year and
optimistic, most likely, and pessimistic salvage values
of $15,000, $10,000, and $2000, respectively,
the salvage value(s) for which the challenger AW
will be lower than that of the defender are:
(a) All of them
(b) Only the optimistic one
(c)
(d)
Only the optimistic and pessimistic ones
None of them
18.49 A decision tree includes all of the following except:
(a) Probability estimates for each outcome
(b) Measure of worth as the selection criterion
(c) Expected results from a decision at each
stage
(d) The MARR
18.50 A real options analysis is most valuable when:
(a) The risk is low and stakes are high
(b) The stakes are low and risk is high
(c) The stakes are high and risk is high
(d) The stakes are low and risk is low
18.51 A small manufacturing company needs to purchase
a machine that will have a first cost of
$70,000. The company wants to buy an option that
will allow it to purchase the machine for the same
price of $70,000 for up to 1 year from now. If the
company’s MARR is 10% per year, the maximum
amount the company should pay for the option is
closest to:
(a) $5850
(b) $6365
(c) $6845
(d) $7295
CASE STUDY
SENSITIVITY TO THE ECONOMIC ENVIRONMENT
Background and Information
Case Study Questions
Berkshire Controllers usually finances its engineering projects
with a combination of debt and equity capital. The resulting
MARR ranges from a low of 4% per year, if business is
slow, to a high of 10% per year. Normally, a 7% per year return
is expected. Also the life estimates for assets tend to go
down about 20% from normal in a vigorous business environment
and up about 10% in a receding economy. The following
estimates are the most likely values for two expansion
plans currently being evaluated. Plan A will be executed at
one location; Plan B will require two locations. All monetary
estimates are in $1000 units.
Plan B
Plan A Location 1 Location 2
First cost, $ 10,000 30,000 5,000
AOC, $ per year 500 100 200
Salvage value, $ 1,000 5,000 200
Estimated life, years 40 40 20
At the weekly meeting, you were asked to examine the following
questions from Berkshire’s president.
1. Are the PW values for plans A and B sensitive to
changes in the MARR?
2. Are the PW values sensitive to varying life estimates?
3. Is the breakeven point for the first cost of plan A sensitive
to the changes in MARR as business goes from vigorous
to receding?

Case Study 511
CASE STUDY
SENSITIVITY ANALYSIS OF PUBLIC SECTOR PROJECTS—WATER SUPPLY PLANS
Background
Information
One of the most basic services provided by municipal governments
is the delivery of a safe, reliable water supply. As
cities grow and extend their boundaries to outlying areas,
they often inherit water systems that were not constructed
according to city codes. The upgrading of these systems is
sometimes more expensive than installing one correctly in
the first place. To avoid these problems, city officials sometimes
install water systems beyond the existing city limits
in anticipation of future growth. This case study was extracted
from such a countywide water and wastewater management
plan and is limited to only some of the water supply
alternatives.
From about a dozen suggested plans, five methods were
developed by an executive committee as alternative ways
of providing water to the study area. These methods were
then subjected to a preliminary evaluation to identify the
most promising alternatives. Six attributes or factors were
used in the initial rating: ability to serve the area, relative
cost, engineering feasibility, institutional issues, environmental
considerations, and lead time requirement. Each
factor carried the same weighting and had values ranging
from 1 to 5, with 5 being best. After the top three alternatives
were identified, each was subjected to a detailed economic
evaluation for selection of the best alternative.
These detailed evaluations included an estimate of the capital
investment of each alternative amortized over 20 years
at 8% per year interest and the annual maintenance and operation
(M&O) costs. The annual cost (an AW value) was
then divided by the population served to arrive at a monthly
cost per household.
Table 18–5 presents the results of the screening using the six
factors rated on a scale of 1 to 5. Alternatives 1A, 3, and 4
were determined to be the three best and were chosen for further
evaluation.
Detailed Cost Estimates
All amounts are cost estimates.
Alternative 1A
Capital cost
Land with water rights: 1720 hectares
@ $5000 per hectare $8,600,000
Primary treatment plant 2,560,000
Booster station at plant 221,425
Reservoir at booster station 50,325
Site cost 40,260
Transmission line from river 3,020,000
Transmission line right-of-way 23,350
Percolation beds 2,093,500
Percolation bed piping 60,400
Production wells 510,000
Well field gathering system 77,000
Distribution system 1,450,000
Additional distribution system 3,784,800
Reservoirs 250,000
Reservoir site, land, and development 17,000
Subtotal 22,758,060
Engineering and contingencies 5,641,940
Total capital investment $28,400,000
TABLE 18–5 Results of Rating Six Factors for Each Alternative, Case Study
Alternative
Description
1A Receive city
water and
recharge wells
3 Joint city and
county plant
4 County treatment
plant
8 Desalt
groundwater
12 Develop military
water
Ability
to
Supply
Area
Relative
Cost
Engineering
Feasibility
Factors
Institutional
Issues
Environmental
Considerations
Lead Time
Requirement Total
5 4 3 4 5 3 24
5 4 4 3 4 3 23
4 4 3 3 4 3 21
1 2 1 1 3 4 12
5 5 4 1 3 1 19

512 Chapter 18 Sensitivity Analysis and Staged Decisions
Maintenance and operation costs (annual)
Pumping 9,812,610 kWh per year
@ $0.08 per kWh $ 785,009
Fixed operating cost 180,520
Variable operating cost 46,730
Taxes for water rights 48,160
Total annual M&O cost $1,060,419
Total annual cost equivalent capital
investment M&O cost
28,400,000( AP ,8%,20)
1,060,419
2,892,540 1,060,419
$3,952,959
Average monthly household cost to serve 95% of 4980
households is
Household cost (3,952,959) ( — 1 ——
$69.63 per month
——
0.95 )
12 ) (
1
4980 ) ( 1
Alternative 3
Total capital investment $29,600,000
Total annual M&O cost $867,119
Total annual cost 29,600,000( AP ,8%,20)
867,119
3,014,760 867,119
$3,881,879
Household cost $68.38 per month
Alternative 4
Total capital investment $29,000,000
Total annual M&O cost $1,063,449
Total annual cost 29,000,000( AP ,8%,20)
1,063,449
2,953,650 1,063,449
$4,017,099
Household cost $70.76 per month
On the basis of the lowest monthly household cost, alternative
3 (joint city and county plant) is the most economically
attractive.
Case Study Exercises
1. If the environmental considerations factor is to have a
weighting of twice as much as any of the other five factors,
what is its percentage weighting?
2. If the ability to supply area and relative cost factors were
each weighted 20% and the other four factors 15% each,
which alternatives would be ranked in the top three?
3. By how much would the capital investment of alternative
4 have to decrease to make it more attractive than
alternative 3?
4. If alternative 1A served 100% of the households instead
of 95%, by how much would the monthly household
cost decrease?
5. (a) Perform a sensitivity analysis on the two parameters
of M&O costs and number of households to
determine if alternative 3 remains the best economic
choice. Three estimates are made for each
parameter in Table 18–6 . M&O costs may vary up
TABLE 18–6 Pessimistic, Most Likely, and Optimistic
Estimates for Two Parameters
Annual
M&O
Costs
Number
of
Households
Alternative 1A
Pessimistic 1% 4980
Most likely $1,060,419 2%
Optimistic −1% 5%
Alternative 3
Pessimistic 5% 4980
Most likely $867,119 2%
Optimistic 0% 5%
Alternative 4
Pessimistic 2% 4980
Most likely $1,063,449 2%
Optimistic −10% 5%

Case Study 513
(b)
(pessimistic) or down (optimistic) from the most
likely estimates presented in the case statement.
The estimated number of households (4980) is determined
to be the pessimistic estimate. Growth of
2% up to 5% (optimistic) will tend to lower the
monthly cost per household.
Consider the monthly cost per household for alternative
4, the optimistic estimate. The number of
households is 5% above 4980, or 5230. What is
the number of households that would have to be
available in order for this option to have exactly
the same monthly household cost as that for alternative
3 at the optimistic estimate of 5230 households?

CHAPTER 19
More on
Variation and
Decision
Making under
Risk
LEARNING OUTCOMES
Purpose: Incorporate decision making under risk into an engineering economy evaluation using probability, sampling,
and simulation.
SECTION TOPIC LEARNING OUTCOME
19.1 Risk versus certainty • Understand the approaches to decision making
under risk and certainty.
19.2 Probability and distributions • Construct a probability distribution and
cumulative distribution for one variable.
19.3 Random sample • Obtain a random sample from a cumulative
distribution using a random number table.
19.4 , , and 2 • Estimate the population expected value, standard
deviation, and variance from a random sample.
19.5 Simulation • Use Monte Carlo sampling and spreadsheetbased
simulation for alternative evaluation.

T
his chapter further expands our ability to analyze variation in estimates, to consider
probability, and to make decisions under risk . Fundamentals discussed include
variables; probability distributions, especially their graphs and properties
of expected value and dispersion; random sampling; and the use of simulation to
account for estimate variation in engineering economy studies.
Through coverage of variation and probability, this chapter complements topics in the first
sections of Chapter 1: the role of engineering economy in decision making and economic
analysis in the problem-solving process. These techniques are more time-consuming than using
estimates made with certainty, so they should be used primarily for critical parameters.
19.1 Interpretation of Certainty,
Risk, and Uncertainty
All things in the world vary—one from another, over time, and with different environments. We
are guaranteed that variation will occur in engineering economy due to its emphasis on decision
making for the future. Except for the use of breakeven analysis, sensitivity analysis, and a very
brief introduction to expected values, virtually all our estimates have been certain; that is, no
variation in the amount has entered into the computations of PW, AW, ROR, or any relations
used. For example, the estimate that cash flow next year will be $4500 is one of certainty. Decision
making under certainty is, of course, not present in the real world now and surely not in the
future. We can observe outcomes with a high degree of certainty, but even this depends upon the
accuracy and precision of the scale or measuring instrument.
To allow a parameter of an engineering economy study to vary implies that risk , and possibly
uncertainty, is introduced.
When there may be two or more observable values for a parameter and it is possible to estimate
the chance that each value may occur, risk is present. Virtually all decision making is
performed under risk .
Risk
As an illustration, decision making under risk is introduced when an annual cash flow estimate
has a 50-50 chance of being either $−1000 or $500.
Decision making under uncertainty means there are two or more values observable, but the
chances of their occurring cannot be estimated or no one is willing to assign the chances. The
observable values in uncertainty analysis are often referred to as states of nature.
For example, consider the states of nature to be the rate of national inflation in a particular
country during the next 2 to 4 years: remain low, increase 2% to 6% annually, or increase 6%
to 8% annually. If there is absolutely no indication that the three values are equally likely, or
that one is more likely than the others, this is a statement that indicates decision making under
uncertainty.
Example 19.1 explains how a parameter can be described and graphed to prepare for decision
making under risk.
EXAMPLE 19.1
CMS in Fairfield, Virginia received three bids each from vendors for two different pieces of
large equipment, A and B. One of each piece of equipment must be purchased. Tom, an engineer
at CMS, performed an evaluation of each bid and assigned it a rating between 0 and 100, with
100 points being the best of the three. The total for each piece of equipment is 100%. The bid
amounts and ratings are shown at the top of Figure 19–1 .
(a) Consider the ratings as the chance out of 100 that the bid will be chosen, and plot cost
versus chance for each vendor.
(b) Since one each of A and B must be purchased, the total cost will vary somewhere between the
sum of the lowest bids ($11 million) and the sum of the highest bids ($25 million). Plot this
range with an equal chance of 1 in 14 that any amount in between these limits is possible.
(c) Discuss the significant difference between the values of the cost ( x axis values) in the
graphs in (a) and (b) above and how the chances are stated ( y axis values).

516 Chapter 19 More on Variation and Decision Making under Risk
Figure 19–1
Plot of cost estimates
versus chance for ( a )
each piece of equipment
and ( b ) total cost
range, Example 19.1.
Equipment A
Equipment B
Bid, $1000 Rating, % Bid, $1000 Rating, %
3,000 65 8,000 33.3
5,000 25 10,000 33.3
10,000 10 15,000 33.3
80
60
65
Equipment A
Chance, %
40
25
20
10
0
1 2 3
4 5
6 7 8 9 10
11 12
Cost, $ million
40
Equipment B
33.3 33.3 33.3
Chance, %
20
0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Cost, $ million
(a) Specific values
—
14 1
Total cost
Chance
0
7 9 11 13 15 17 19 21 23 25 27 29
Cost, $ million
(b) Continuous range
Solution
(a) Figure 19–1 a plots the specific bids for equipment A and B. The chances (ratings) for A and
for B add to 100%. No values between the specific bids have any chance of occurring, according
to the single-estimate bids from the three vendors.
(b) The range of total cost is between $11 million and $25 million, as shown in Figure 19–1 b .
Tom decided to make his estimate of total cost continuous between these two extremes.
This means that the discrete sums of bids ($11 million, $15 million, and $25 million) are
no longer used. Rather the entire range from $11 million to $25 million with a chance for
every total cost in between is included. Every value has a chance of 1 in 14 of being observed.
Now, the sum is a continuous value.
(c) In the graph for bid values (Figure 19–1 a ), only specific or discrete estimates are included
on the x axis. In the graph for the sum of the cost for equipment A and B (Figure 19–1 b ),
the y axis values are continuous over a specific range.
In the next section, the term variable is defined and two types of variables are explained—
discrete and continuous —as illustrated here in an elementary form.

19.1 Interpretation of Certainty, Risk, and Uncertainty 517
Before initiating an engineering economy study, it is important to decide if the analysis will
be conducted with certainty for all parameters or if risk will be introduced. A summary of the
meaning and use for each type of analysis follows.
Decision Making under Certainty This is what we have done in most analyses thus far.
Deterministic estimates are made and entered into measure of worth relations—PW, AW, FW,
ROR, BC—and decision making is based on the results. The values estimated can be considered
the most likely to occur with all chance placed on the single-value estimate. A typical example is
an asset’s first cost estimate made with certainty, say, P $50,000. A plot of P versus chance has
the general form of Figure 19–1 a with one vertical bar at $50,000 and 100% chance placed on it.
The term deterministic , in lieu of certainty , is often used when single-value or single-point
estimates are used exclusively.
In fact, sensitivity analysis using different values of an estimate is simply another form of
analysis with certainty, except that the analysis is repeated with different values, each estimated
with certainty. The resulting measure of worth values are calculated and graphically
portrayed to determine the decision’s sensitivity to different estimates for one or more
parameters.
Decision Making under Risk Now the element of chance is formally taken into account.
However, it is more difficult to make a clear decision because the analysis attempts to accommodate
variation . One or more parameters in an alternative will be allowed to vary. The estimates
will be expressed as in Example 19.1 or in slightly more complex forms. Fundamentally,
there are two ways to consider risk in an analysis:
Expected value analysis. Use the chance and parameter estimates to calculate expected
values E (parameter) via formulas such as Equation [18.2]. Analysis results in E (cash flow),
E (AOC), and the like; and the final result is the expected value for a measure of worth, such
as E (PW), E (AW), E (ROR), E (BC). To select the alternative, choose the most favorable
expected value of the measure of worth. In an elementary form, this is what we learned about
expected values in Chapter 18. The computations may become more elaborate, but the principle
is fundamentally the same.
Simulation analysis. Use the chance and parameter estimates to generate repeated computations
of the measure of worth relation by randomly sampling from a plot for each varying
parameter similar to those in Figure 19–1 . When a representative and random sample is complete,
an alternative is selected utilizing a table or plot of the results. Usually, graphics are an
important part of decision making via simulation analysis. Basically, this is the approach discussed
in the rest of this chapter.
Decision Making under Uncertainty When chances are not known for the identified
states of nature (or values) of the uncertain parameters, the use of expected value–based decision
making under risk as outlined above is not an option . In fact, it is difficult to determine
what criterion to use to even make the decision. If it is possible to agree that each state is
equally likely, then all states have the same chance, and the situation reduces to one of decision
making under risk, because expected values can be determined. Because of the relatively
inconclusive approaches necessary to incorporate decision making under uncertainty into an
engineering economy study, the techniques can be quite useful but are beyond the intended
scope of this text.
In an engineering economy study, observed parameter values will vary from the value estimated
at the time of the study. However, when performing the analysis, not all parameters should
be considered as probabilistic (or at risk). Those that are estimable with a relatively high degree
of certainty should be fixed for the study. Accordingly, the methods of sampling, simulation, and
statistical data analysis are selectively used on parameters deemed important to the decisionmaking
process. Parameters such as P , AOC, material and unit costs, sales price, revenues, etc.,
are the targets of decision making under risk. Anticipated variation in interest rates is more commonly
addressed by sensitivity analysis.
The remainder of this chapter concentrates on decision making under risk as applied in an
engineering economy study. Sections 19.2 to 19.4 provide foundation material necessary to design
and correctly conduct a simulation analysis (Section 19.5).

518 Chapter 19 More on Variation and Decision Making under Risk
Figure 19–2
( a ) Discrete and continuous
variable scales and
( b) scales for a variable
versus its probability.
Discrete
variable
Continuous
variable
1.0
3 5 10 15 Years
Estimated life, n
…
–100%
∞
Rate of return, i
(a)
P(n)
P(i)
0.8
0.6
0.4
0.2
0
1.0
0.8
0.6
0.4
0.2
Discrete variable
Estimated life,
probability vs. years
3 5 10 15 n, years
Continuous variable
Rate of return,
probability vs. percent
10 5 0 5 10 15 20 25 30 35 … i, %
(b)
19.2 Elements Important to Decision
Making under Risk
Some basics of probability and statistics are essential to correctly perform decision making under
risk via expected value or simulation analysis. They are the random variable, probability, probability
distribution , and cumulative distribution , as defined here. (If you are already familiar with
them, this section will provide a review.)
A random variable or variable is a characteristic or parameter that can take on any one of several
values. Variables are classified as discrete or continuous. Discrete variables have several
specific, isolated values, while continuous variables can assume any value between two stated
limits, called the range of the variable.
The estimated life of an asset is a discrete variable. For example, n may be expected to
have values of n 3, 5, 10, or 15 years, and no others. The rate of return is an example of a
continuous variable; i can vary from −100% to , that is, −100% i . The ranges of possible
values for n (discrete) and i (continuous) are shown as the x axes in Figure 19–2 a . (In
probability texts, capital letters symbolize a variable, say X , and small letters x identify a
specific value of the variable. Though correct, this level of rigor in terminology is not applied
in this chapter.)
Probability is a number between 0 and 1.0 that expresses the chance in decimal form that a random
variable (discrete or continuous) will take on any value from those identified for it. Probability
is simply the amount of chance, divided by 100.
Probabilities are commonly identified by P ( X i ) or P ( X X i ), which is read as the probability that
the variable X takes on the value X i . (Actually, for a continuous variable, the probability at a
single value is zero, as shown in a later example.) The sum of all P ( X i ) for a variable must be 1.0,

19.2 Elements Important to Decision Making under Risk 519
a requirement already discussed. The probability scale, like the percentage scale for chance in
Figure 19–1 , is indicated on the ordinate ( y axis) of a graph. Figure 19–2 b shows the 0 to 1.0
range of probability for the variables n and i.
A probability distribution describes how probability is distributed over the different values of
a variable. Discrete variable distributions look significantly different from continuous variable
distributions, as indicated by the inset at the right.
The individual probability values are stated as
P ( X i ) probability that X equals X i [19.1]
The distribution may be developed in one of two ways: by listing each probability value for each
possible variable value (see Example 19.2) or by a mathematical description or expression that
states probability in terms of the possible variable values (Example 19.3).
Cumulative distribution, also called the cumulative probability distribution , is the accumulation
of probability over all values of a variable up to and including a specified value.
Identified by F ( X i ), each cumulative value is calculated as
P(X i )
P(X i )
F(X i )
Discrete
X i
Continuous
X i
Discrete
F ( X i ) sum of all probabilities through the value X i
P ( X X i ) [19.2]
As with a probability distribution, cumulative distributions appear differently for discrete (stairstepped)
and continuous variables (smooth curve). Examples 19.2 and 19.3 illustrate cumulative
distributions that correspond to specific probability distributions. These fundamentals about
F ( X i ) are applied in the next section to develop a random sample.
EXAMPLE 19.2
Alvin is a medical doctor and biomedical engineering graduate who practices at Medical Center
Hospital. He is planning to start prescribing an antibiotic that may reduce infection in patients
with flesh wounds. Tests indicate the drug has been applied up to 6 times per day without
harmful side effects. If no drug is used, there is always a positive probability that the infection
will be reduced by a person’s own immune system.
Published drug test results provide good probability estimates of positive reaction (i.e., reduction
in the infection count) within 48 hours for increased treatments per day. Use the probabilities
listed below to construct a probability distribution and a cumulative distribution for
the total number of treatments per day.
Number of Added
Treatments per Day
Probability of Infection Reduction
for Each Added Treatment
0 0.07
1 0.08
2 0.10
3 0.12
4 0.13
5 0.25
6 0.25
Solution
Define the random variable T as the number of added treatments per day. Since T can take on
only seven different values, it is a discrete variable . The probability of infection reduction is
listed for each value in column 2 of Table 19–1 . The cumulative probability F ( T i ) is determined
using Equation [19.2] by adding all P ( T i ) values through T i , as indicated in column 3.
Figure 19–3 a and b shows plots of the probability distribution and cumulative distribution,
respectively. The summing of probabilities to obtain F ( T i ) gives the cumulative distribution the
stair-stepped appearance, and in all cases the final F ( T i ) 1.0, since the total of all P ( T i ) values
must equal 1.0.
F(X i )
Continuous
X i
X i

520 Chapter 19 More on Variation and Decision Making under Risk
TABLE 19–1 Probability Distribution and Cumulative
Distribution for Example 19.2
(1) (2) (3)
Number per Day
T i
Probability
P ( T i )
Cumulative
Probability
F ( T i )
0 0.07 0.07
1 0.08 0.15
2 0.10 0.25
3 0.12 0.37
4 0.13 0.50
5 0.25 0.75
6 0.25 1.00
Figure 19–3
( a ) Probability distribution
P ( T i ) and
( b ) cumulative distribution
F ( T i ) for
Example 19.2.
P(T i )
0.3
0.25 0.25
0.2
0.1
0.07 0.08 0.10 0.12 0.13
F(T i )
1.0
0.8
0.6
0.3
0.15
0.07
0.25
0.37
0.50
0.75
1.00
0
0 1 2 3 4 5 6 T i
(a)
0
0 1 2 3 4 5 6 T i
(b)
Comment
Rather than use a tabular form as in Table 19–1 to state P ( T i ) and F ( T i ) values, it is possible to
express them for each value of the variable.
{
0.07 T 1 0
0.08 T 2 1
0.10 T 3
{
0.07 T 1 0
0.15 T 2 1
2
0.25 T 3 2
P(T i ) 0.12 T 4 3 F(T i ) 0.37 T 4 3
0.13 T 5 4
0.50 T 5 4
0.25 T 6 5
0.75 T 6 5
0.25 T 7 6
1.00 T 7 6
In basic engineering economy situations, the probability distribution for a continuous variable
is commonly expressed as a mathematical function, such as a uniform distribution, a triangular
distribution (both discussed in Example 19.3 in terms of cash flow), or the more complex, but
commonly used, normal distribution. For continuous variable distributions, the symbol f (X ) is
routinely used instead of P (X i ), and F (X ) is used instead of F (X i ), simply because the point probability
for a continuous variable is zero. Thus, f (X ) and F (X ) are continuous lines and curves.
EXAMPLE 19.3
As president of a manufacturing systems consultancy, Sallie has observed the monthly cash
flows that have occurred over the last 3 years into company accounts from two longstanding
clients. Sallie has concluded the following about the distribution of these monthly cash flows:

19.2 Elements Important to Decision Making under Risk 521
Client 1 Client 2
Estimated low cash flow: $10,000 Estimated low cash flow: $20,000
Estimated high cash flow: $15,000 Estimated high cash flow: $30,000
Most likely cash flow: same for all values Most likely cash flow: $28,000
Distribution of probability: uniform Distribution of probability: mode at $28,000
The mode is the most frequently observed value for a variable. Sallie assumes cash flow to be
a continuous variable referred to as C . ( a ) Write and graph the two probability distributions and
cumulative distributions for monthly cash flow, and ( b ) determine the probability that monthly
cash flow is no more than $12,000 for client 1 and at least $25,000 for client 2.
Solution
All cash flow values are expressed in $1000 units.
Client 1: monthly cash fl ow distribution
(a) The distribution of cash flows for client 1, identified by the variable C 1 , follows the uniform
distribution. Probability and cumulative probability take the following general forms.
f ( C 1 ) ————— 1
high low
low value C 1 high value
f ( C 1 ) ——— 1
H L
L C 1 H [19.3]
F ( C 1 ) ——————
value low
low value C
high low
1 high value
F ( C 1 ) C 1
———
L
H L
L C 1 H [19.4]
For client 1, monthly cash flow is uniformly distributed with L $10, H $15, and $10
C 1 $15. Figure 19–4 is a plot of f ( C 1 ) and F ( C 1 ) from Equations [19.3] and [19.4].
f ( C 1 ) 1 —
5 0.2 $10 C 1 $15
F ( C 1 ) C 1 10 ————
5
$10 C 1 $15
(b) The probability that client 1 has a monthly cash flow of no more than $12 is easily determined
from the F ( C 1 ) plot as 0.4, or a 40% chance. If the F ( C 1 ) relation is used directly, the
computation is
F ($12) P ( C 1 $12) ———— 12 10
0.4
5
1.0
F(C 1 )
Figure 19–4
Uniform distribution for
client 1 monthly cash
flow, Example 19.3.
0.8
0.6
0.4
f (C 1 )
0.2
0.2
0
10 15
0
10
12 15
C 1 C 1

522 Chapter 19 More on Variation and Decision Making under Risk
Client 2: monthly cash fl ow distribution
(a) The distribution of cash flows for client 2, identified by the variable C 2 , follows the triangular
distribution. This probability distribution has the shape of an upward-pointing triangle
with the peak at the mode M , and downward-sloping lines joining the x axis on either
side at the low ( L ) and high ( H ) values. The mode of the triangular distribution has the
maximum probability value.
f (mode) f ( M ) ——— 2
[19.5]
H L
The cumulative distribution is comprised of two curved line segments from 0 to 1 with a
break point at the mode, where
F (mode) F ( M ) ——— M H L
L
[19.6]
For C 2 , the low value is L $20, the high is H $30, and the most likely cash flow is the
mode M $28. The probability at M from Equation [19.5] is
f (28) ———— 2
30 20 ——
10 2 0.2
The break point in the cumulative distribution occurs at C 2 28. Using Equation [19.6],
F (28) ———— 28 20
30 20 0.8
Figure 19–5 presents the plots for f ( C 2 ) and F ( C 2 ). Note that f ( C 2 ) is skewed, since the
mode is not at the midpoint of the range H L , and F ( C 2 ) is a smooth S-shaped curve with
an inflection point at the mode.
(b) From the cumulative distribution in Figure 19–5 , there is an estimated 31.25% chance that
cash flow is $25 or less. Therefore,
F ($30) F ($25) P ( C 2 $25) 1 0.3125 0.6875
Comment
The general relations f ( C 2 ) and F ( C 2 ) are not developed here. The variable C 2 is not a uniform
distribution; it is triangular. Therefore, it requires the use of an integral to find cumulative probability
values from the probability distribution f ( C 2 ).
F(C 2 )
1.0
Mode
0.8
0.6
0.2
f (C 2 )
Mode
0.4
0.3125
0.2
0
20 28 30
0
20 25 28 30
C 2 C 2
Figure 19–5
Triangular distribution for client 2 monthly cash flow, Example 19.3.

19.3 Random Samples 523
19.3 Random Samples
Estimating a parameter with a single value in previous chapters is the equivalent of taking a random
sample of size 1 from an entire population of possible values. As an illustration, assume that
estimates of first cost, annual operating cost, interest rate, and other parameters are used to compute
one PW value in order to accept or reject an alternative. Each estimate is a sample of size 1
from an entire population of possible values for each parameter. Now, if a second estimate is
made for each parameter and a second PW value is determined, a sample of size 2 has been taken.
If all values in the population were known, the probability distribution and cumulative distribution
would be known. Then a sample would not be necessary.
When we perform an engineering economy study and utilize decision making under certainty,
we use one estimate for each parameter to calculate a measure of worth (i.e., a sample of size 1
for each parameter). The estimate is the most likely value, that is, one estimate of the expected
value. We know that all parameters will vary somewhat; yet some are important enough, or will
vary enough, that a probability distribution should be determined or assumed for it and the parameter
treated as a random variable. This is using risk, and a sample from the parameter’s probability
distribution— P ( X ) for discrete or f ( X ) for continuous—helps formulate probability statements
about the estimates. This approach complicates the analysis somewhat; however, it also
provides a sense of confidence (or possibly a lack of confidence in some cases) about the decision
made concerning the economic viability of the alternative based on the varying parameter. (We
will further discuss this aspect later, after we learn how to correctly take a random sample from
any probability distribution.)
A random sample of size n is the selection in a random fashion of n values from a population
with an assumed or known probability distribution, such that the values of the variable have the
same chance of occurring in the sample as they are expected to occur in the population.
Suppose Yvon is an engineer with 20 years of experience working for the Aircraft Safety
Commission. For a two-crew aircraft, there are three parachutes on board. The safety standard
states that 99% of the time, all three chutes must be “fully ready for emergency deployment.”
Yvon is relatively sure that nationwide the probability distribution of N , the specific number of
chutes fully ready, may be described by the probability distribution
{
0.005 N 0 chutes ready
0.015 N 1 chute ready
P(N N i )
0.060 N 2 chutes ready
0.920 N 3 chutes ready
This means that the safety standard is clearly not met nationwide. Yvon is in the process of
sampling 200 (randomly selected) corporate and private aircraft across the nation to determine
how many chutes are classified as fully ready. If the sample is truly random and Yvon’s
probability distribution is a correct representation of actual parachute readiness, the observed
N values in the 200 aircraft will approximate the same proportions as the population
probabilities, that is, 1 aircraft with 0 chutes ready, etc. Since this is a sample, it is likely that
the results won’t track the population exactly. However, if the results are relatively close, the
study indicates that the sample results may be useful in predicting parachute safety across
the nation.
To develop a random sample, use random numbers ( RN ) generated from a uniform probability
distribution for the discrete numbers 0 through 9, that is,
P ( X i ) 0.1 for X i 0, 1, 2, … , 9
In tabular form, the random digits so generated are commonly clustered in groups of two digits,
three digits, or more. Table 19–2 is a sample of 264 random digits clustered into two-digit numbers.
This format is very useful because the numbers 00 to 99 conveniently relate to the cumulative
distribution values 0.01 to 1.00. This makes it easy to select a two-digit RN and enter F ( X )
to determine a value of the variable with the same proportions as it occurs in the probability distribution.
To apply this logic manually and develop a random sample of size n from a known

524 Chapter 19 More on Variation and Decision Making under Risk
TABLE 19–2
Random Digits Clustered into Two-Digit Numbers
51 82 88 18 19 81 03 88 91 46 39 19 28 94 70 76 33 15 64 20 14 52
73 48 28 59 78 38 54 54 93 32 70 60 78 64 92 40 72 71 77 56 39 27
10 42 18 31 23 80 80 26 74 71 03 90 55 61 61 28 41 49 00 79 96 78
45 44 79 29 81 58 66 70 24 82 91 94 42 10 61 60 79 30 01 26 31 42
68 65 26 71 44 37 93 94 93 72 84 39 77 01 97 74 17 19 46 61 49 67
75 52 14 99 67 74 06 50 97 46 27 88 10 10 70 66 22 56 18 32 06 24
discrete probability distribution P ( X ) or a continuous variable distribution f ( X ), the following
procedure may be used.
1. Develop the cumulative distribution F ( X ) from the probability distribution. Plot F ( X ).
2. Assign the RN values from 00 to 99 to the F ( X ) scale (the y axis) in the same proportion as
the probabilities. For the parachute safety example, the probabilities from 0.0 to 0.15 are
represented by the random numbers 00 to 14. Indicate the RNs on the graph.
3. To use a table of random numbers, determine the scheme or sequence of selecting RN
values—down, up, across, diagonally. Any direction and pattern is acceptable, but the
scheme should be used consistently for one entire sample.
4. Select the first number from the RN table, enter the F ( X ) scale, and observe and record the
corresponding variable value. Repeat this step until there are n values of the variable that
constitute the random sample.
5. Use the n sample values for analysis and decision making under risk. These may include
• Plotting the sample probability distribution.
• Developing probability statements about the parameter.
• Comparing sample results with the assumed population distribution.
• Determining sample statistics (Section 19.4).
• Performing a simulation analysis (Section 19.5).
EXAMPLE 19.4
Develop a random sample of size 10 for the variable N , number of months, as described by the
probability distribution
0.20 N 24
P(N N i ) 0.50 N 30
[19.7]
{
0.30 N 36
Solution
Apply the procedure above, using the P ( N N i ) values in Equation [19.7].
1. The cumulative distribution, Figure 19–6 , is for the discrete variable N , which can assume
three different values.
2. Assign 20 numbers (00 through 19) to N 1 24 months, where P ( N 24) 0.2; 50 numbers
to N 2 30; and 30 numbers to N 3 36.
3. Initially select any position in Table 19–2 , and go across the row to the right and onto the
row below toward the left. (Any routine can be developed, and a different sequence for
each random sample may be used.)
4. Select the initial number 45 (4th row, 1st column), and enter Figure 19–6 in the RN range
of 20 to 69 to obtain N 30 months.
5. Select and record the remaining nine values from Table 19–2 as shown below.
RN 45 44 79 29 81 58 66 70 24 82
N 30 30 36 30 36 30 30 36 30 36

19.3 Random Samples 525
Cumulative
probability
1.0
0.7
Assigned
random
numbers
70–99
Figure 19–6
Cumulative distribution
with random number values
assigned in proportion
to probabilities,
Example 19.4.
F(N i )
20–69
45
0.2
00–19
24 30 36
N i , months
Now, using the 10 values, develop the sample probabilities.
Months
N
Times in
Sample
Sample
Probability
Equation [19.7]
Probability
24 0 0.00 0.2
30 6 0.60 0.5
36 4 0.40 0.3
With only 10 values, we can expect the sample probability estimates to be different from the
values in Equation [19.7]. Only the value N 24 months is significantly different, since no RN
of 19 or less occurred. A larger sample will definitely make the probabilities closer to the
original data.
To take a random sample of size n for a continuous variable, the procedure above is applied,
except the random number values are assigned to the cumulative distribution on a continuous
scale of 00 to 99 corresponding to the F ( X ) values. As an illustration, consider Figure 19–4 ,
where C 1 is the uniformly distributed cash flow variable for client 1 in Example 19.3. Here L
$10, H $15, and f ( C 1 ) 0.2 for all values between L and H (all values are divided by $1000).
The F ( C 1 ) is repeated as Figure 19–7 with the assigned random number values shown on the right
scale. If the two-digit RN of 45 is chosen, the corresponding C 1 is graphically estimated to be
$12.25. It can also be linearly interpolated as $12.25 10 (45100)(15 – 10).
Cumulative
probability
1.0
0.8
Assigned
random
numbers
99
79
Figure 19–7
Random numbers
assigned to the
continuous variable of
client 1 cash flows in
Example 19.3.
F(C 1 )
0.6
0.4
59
39
45
0.2
19
00
10 11 12 13 14 15
12.25
C 1 , $1000

526 Chapter 19 More on Variation and Decision Making under Risk
For greater accuracy when developing a random sample, especially for a continuous variable,
it is possible to use 3-, 4-, or 5-digit RNs. These can be developed from Table 19–2 simply by
combining digits in the columns and rows or obtained from tables with RNs printed in larger
clusters of digits. In computer-based sampling, most simulation software packages have an RN
generator built in that will generate values in the range of 0 to 1 from a continuous variable uniform
distribution, usually identified by the symbol U (0, 1). The RN values, usually between
0.00000 and 0.99999, are used to sample directly from the cumulative distribution employing
essentially the same procedure explained here. The Excel functions RAND and RANDBE-
TWEEN are described in Appendix A, Section A.3.
An initial question in random sampling usually concerns the minimum size of n required to
ensure confidence in the results. Without detailing the mathematical logic, sampling theory,
which is based upon the law of large numbers and the central limit theorem (check a basic statistics
book to learn about these), indicates that an n of 30 is sufficient. However, since reality does
not follow theory exactly, and since engineering economy often deals with sketchy estimates,
samples in the range of 100 to 200 are the common practice. But samples as small as 10 to 25
provide a much better foundation for decision making under risk than the single-point estimate
for a parameter that is known to vary widely.
19.4 Expected Value and Standard Deviation
Two very important measures or properties of a random variable are the expected value and standard
deviation. If the entire population for a variable were known, these properties would be
calculated directly. Since they are usually not known, random samples are commonly used to
estimate them via the sample mean and the sample standard deviation, respectively. The following
is a brief introduction to the interpretation and calculation of these properties using a random
sample of size n from the population.
The usual symbols are Greek letters for the true population measures and English letters for
the sample estimates.
True Population Measure
Sample Estimate
Symbol Name Symbol Name
Expected value or E ( X ) Mu or true mean
__
X Sample mean
Standard deviation
______
__
or √Var(X)
___ Sigma or true s or √s Sample standard
or √ standard deviation
deviation
The expected value E(X) is the long-run expected average if the variable is sampled many times.
The population expected value is not known exactly, since the population itself is not known
completely, so µ is estimated either by E ( X ) from a distribution or by X __
, the sample mean. Equation
[18.2], repeated here as Equation [19.8], is used to compute the E ( X ) of a probability distribution,
and Equation [19.9] is the sample mean, also called the sample average .
Population:
Probability distribution: E(X) X i P(X i ) [19.8]
Sample:
__ sum of sample values
X ——————————
sample size
X i
——
n f i X i
———
n [19.9]
The f i in the second form of Equation [19.9] is the frequency of X i , that is, the number of times
each value occurs in the sample. The resulting X __
is not necessarily an observed value of the variable;
it is the long-run average value and can take on any value within the range of the variable.
(We omit the subscript i on X and f when there is no confusion introduced.)

19.4 Expected Value and Standard Deviation 527
EXAMPLE 19.5
Kayeu, an engineer with Pacific NW Utilities, is planning to test several hypotheses about
residential electricity bills in North American and Asian countries. The variable of interest is
X , the monthly residential bill in U.S. dollars (rounded to the nearest dollar). Two small samples
have been collected from different countries of North America and Asia. Estimate the
population expected value. Do the samples (from a nonstatistical viewpoint) appear to be
drawn from one population of electricity bills or from two different populations?
North American, Sample 1, $ 40 66 75 92 107 159 275
Asian, Sample 2, $ 84 90 104 187 190
Solution
Use Equation [19.9] for the sample mean.
Sample 1: n 7 X i 814
Sample 2: n 5 X i 655
__
X __ $116.29
X $131.00
Based solely on the small sample averages, the approximate $15 difference, which is
only11% of the larger average bill, does not seem sufficiently large to conclude that the two
populations are different. There are several statistical tests available to determine if samples
come from the same or different populations. (Check a basic statistics text to learn
about them.)
Comment
There are three commonly used measures of central tendency for data. The sample average
is the most popular, but the mode and the median are also good measures. The mode, which
is the most frequently observed value, was utilized in Example 19.3 for a triangular distribution.
There is no specific mode in Kayeu’s two samples, since all values are different. The
median is the middle value of the sample. It is not biased by extreme sample values, as is the
mean. The two medians in the samples are $92 and $104. Based solely on the medians, the
conclusion is still that the samples do not necessarily come from two different populations
of electricity bills.
The standard deviation s or s(X ) is the dispersion or spread of values about the expected value
E(X ) or sample average X __
.
The sample standard deviation s estimates the property , which is the population measure of
dispersion about the expected value of the variable. A probability distribution for data with strong
central tendency is more closely clustered about the center of the data, and has a smaller s , than
a wider, more dispersed distribution. In Figure 19–8 , the samples with larger s values— s 1 and
s 4 —have a flatter, wider probability distribution.
f(X)
P(X)
s 2
s 1
s 3
s 4
X 1
X 2
s 1 s 2 s 4 s 3
Figure 19–8
Sketches of distributions with different separate lines standard deviation values.

528 Chapter 19 More on Variation and Decision Making under Risk
EXAMPLE 19.6
Actually, the variance s 2 is often quoted as the measure of dispersion. The standard deviation
is simply the square root of the variance, so either measure can be used. The s value is what we
use routinely in making computations about risk and probability. Mathematically, the formulas
and symbols for variance and standard deviation of a discrete variable and a random sample of
size n are as follows:
___
Population: 2 Var ( X ) and √ 2 √ _______
Var(X)
Probability distribution: Var ( X ) [X i E(X)] 2 P(X i ) [19.10]
Sample: s 2 sum of (sample value sample average)2
——————————————————
sample size 1
(X i X __
) 2
—————
[19.11]
n 1
__
s √s 2
Equation [19.11] for sample variance is usually applied in a more computationally convenient
form.
s 2 X i 2
———
n 1
__
——— n 2
X f i X i 2
———
n 1 n 1
__
——— n 2
X [19.12]
n 1
The standard deviation uses the sample average as a basis about which to measure the spread or
dispersion of data via the calculation ( X X __
), which can have a minus or plus sign. To accurately
measure the dispersion in both directions from the average, the quantity ( X X __
) is squared. To
return to the dimension of the variable itself, the square root of Equation [19.11] is extracted. The
term ( X X __
) 2 is called the mean-squared deviation, and s has historically also been referred to
as the root-mean-square deviation. The f i in the second form of Equation [19.12] uses the frequency
of each X i value to calculate s 2 .
One simple way to combine the average and standard deviation is to determine the percentage
or fraction of the sample that is within 1, 2, or 3 standard deviations of the average, that is,
__
X ts for t 1, 2, or 3 [19.13]
In probability terms, this is stated as
P ( __
X ts X __
X ts ) [19.14]
Virtually all the sample values will always be within the 3 s range of __
__ X , but the percent within
1 s will vary depending on how the data points are distributed about X . Example 19.6 illustrates
the calculation of s to estimate and incorporates s with the sample average using X __
ts.
( a ) Use the two samples of Example 19.5 to estimate population variance and standard deviation
for electricity bills. ( b ) Determine the percentages of each sample that are inside the ranges
of 1 and 2 standard deviations from the mean.
Solution
(a) For illustration purposes only, apply the two different relations to calculate s for the two
samples. For sample 1 (North American) with n 7, use X to identify the values. Table 19–3
presents the computation of (X X __
) 2 for Equation [19.11], with X __
$116.29. The resulting
s 2 and s values are
s 2 —————
37,743.40 6290.57
6
s $79.31

19.4 Expected Value and Standard Deviation 529
TABLE 19–3 Computation of Standard Deviation
Using Equation [19.11] with X __
$116.29, Example 19.6
X , $
X − X
__
( X − X __
) 2
40 76.29 5,820.16
66 50.29 2,529.08
75 41.29 1,704.86
92 24.29 590.00
107 9.29 86.30
159 42.71 1,824.14
275 158.71 25,188.86
814 37,743.40
TABLE 19–4 Computation of Standard Deviation
Using Equation [19.12] with __
Y $131,
Example 19.6
For sample 2 (Asian), use Y to identify the values. With n 5 and __
Y 131, Table 19–4
shows Y 2 for Equation [19.12]. Then
s 2 ———
97,041
4
s $53
Y , $ Y 2
84 7,056
90 8,100
104 10,816
187 34,969
190 36,100
655 97,041
5 —
4 (131) 2 42,260.25 1.25(17,161) 2809
The dispersion is smaller for the Asian sample ($53) than for the North American sample
($79.31).
(b) Equation [19.13] determines the ranges of X __
1 s and X __
2 s . Count the number of sample
data points between the limits, and calculate the corresponding percentage. See Figure 19–9
for a plot of the data and the standard deviation ranges.
–
X = 116.29
– 80 – 40 0 40 80 120
–
X ± 1s
–
X ± 2s
(a)
–
Y = 131
160 200 240 280 X
0 40 80 120
–
Y ± 1s
–
Y ± 2s
Figure 19–9
Values, averages, and standard deviation ranges for ( a ) North American and
( b ) Asian samples, Example 19.6.
(b)
160 200 240 Y

530 Chapter 19 More on Variation and Decision Making under Risk
North American sample
__
X 1 s 116.29 79.31 for a range of $36.98 to $195.60
Six out of seven values are within this range, so the percentage is 85.7%.
__
X 2 s 116.29 158.62 for a range of $42.33 to $274.91
There are still six of the seven values within the X __
2 s range. The limit $42.33 is meaningful
only from the probabilistic perspective; from the practical viewpoint, use zero, that
is, no amount billed.
Asian sample
__
Y 1 s 131 53 for a range of $78 to $184
There are three of five values, or 60%, within the range.
__
Y 2 s 131 106 for a range of $25 to $237
All five of the values are within the Y __
2 s range.
Comment
A second common measure of dispersion is the range, which is simply the largest minus the
smallest sample values. In the two samples here, the range estimates are $235 and $106.
Only the hand computations for E ( X ), s , and s 2 have been demonstrated here. Calculators and
spreadsheets all have functions to determine these values by simply entering the data.
Before we perform simulation analysis in engineering economy, it may be of use to summarize
the expected value and standard deviation relations for a continuous variable, since Equations
[19.8] through [19.12] address only discrete variables. The primary differences are that the
summation symbol is replaced by the integral over the defined range of the variable, which we
identify as R , and that P ( X ) is replaced by the differential element f ( X ) dX . For a stated continuous
probability distribution f ( X ), the formulas are
Expected value: E ( X ) R
Xf ( X ) dX [19.15]
Variance: Var ( X ) R
X 2 f ( X ) dX [ E(X)] 2 [19.16]
For a numerical example, again use the uniform distribution in Example 19.3 ( Figure 19–4 ) over
the range R from $10 to $15. If we identify the variable as X , rather than C 1 , the following are
correct.
f ( X ) — 1 0.2 $10 X $15
5
E ( X ) X(0.2) dX 0.1X2 R | 15 0.1(225 100) $12.5
10
Var( X ) X2 (0.2) dX (12.5) 2 —— 0.2
R 3 | 15 X3 10 (12.5) 2
0.06667(3375 1000) 156.25 2.08
____
√2.08 $1.44
Therefore, the uniform distribution between L $10 and H $15 has an expected value of
$12.5 (the midpoint of the range, as expected) and a standard deviation of $1.44.
EXAMPLE 19.7
Christy is the regional safety engineer for a chain of franchise-based gasoline and food stores.
The home office has had many complaints and several legal actions from employees and customers
about slips and falls due to liquids (water, oil, gas, soda, etc.) on concrete surfaces.

19.4 Expected Value and Standard Deviation 531
Corporate management has authorized each regional engineer to contract locally to apply to all
exterior concrete surfaces a newly marketed product that absorbs up to 100 times its own
weight in liquid and to charge a home office account for the installation. The authorizing letter
to Christy states that, based upon their simulation and random samples that assume a normal
population, the cost of the locally arranged installation should be about $10,000 and almost
always is within the range of $8000 to $12,000.
You have been asked to write a brief but thorough summary about the normal distribution,
explain the $8000 to $12,000 range statement, and explain the phrase “random samples that
assume a normal population.”
Solution
The following summaries about the normal distribution and sampling will help explain the
authorization letter.
Normal distribution, probabilities, and random samples
The normal distribution is also referred to as the bell-shaped curve , the Gaussian distribution
, or the error distribution . It is, by far, the most commonly used probability distribution
in all applications. It places exactly one-half of the probability on either side of the mean or
expected value. It is used for continuous variables over the entire range of numbers. The
normal distribution is found to accurately predict many types of outcomes, such as IQ values;
manufacturing errors about a specified size, volume, weight, etc.; and the distribution of
sales revenues, costs, and many other business parameters around a specified mean, which is
why it may apply in this situation.
The normal distribution, identified by the symbol N ( , 2 ), where is the expected value or
mean and 2 is the variance, or measure of spread, can be described as follows:
• The mean locates the probability distribution ( Figure 19–10 a ), and the spread of the distribution
varies with variance ( Figure 19–10 b ), growing wider and flatter for larger variance
values.
• When a sample is taken, the estimates are identified as sample mean X __
for and sample
standard deviation s for .
• The normal probability distribution f ( X ) for a variable X is quite complicated, because its
formula is
f ( X ) ——— 1___
√2
exp { [
where exp represents the number e 2.71828.
(X )2
————
2 2 ] }
Since f ( X ) is so unwieldy, random samples and probability statements are developed using a
transformation, called the standard normal distribution (SND), which uses and (population)
or X __
and s (sample) to compute values of the variable Z .
Population :
Sample:
Z —————————
deviation from mean
standard deviation
———
X
[19.17]
Z X X
__
———
s [19.18]
The SND for Z ( Figure 19–10 c ) is the same as for X , except that it always has a mean of 0 and
a standard deviation of 1, and it is identified by the symbol N (0, 1). Therefore, the probability
values under the SND curve can be stated exactly. It is always possible to transfer back to the
original values from sample data by solving Equation [19.17] for X :
X Z [19.19]

532 Chapter 19 More on Variation and Decision Making under Risk
Figure 19–10
Normal distribution
showing ( a ) different
mean values µ , ( b )
different standard deviation
values , and
( c ) relation of
normal X to standard
normal Z .
f(X)
1
2 3
1 2 3
(a)
X
• Equal dispersion
( 1 = 2 = 3 )
• Increasing means
f(X)
1
• Increasing
dispersion
( 1 < 2 < 3 )
• Two different
means
3
1 = 3 2
2
(b)
X
f(X)
Normal distribution
– 3
+ 3
f(Z)
0.9974
0.9546
0.6826
X
0.3413 0.3413
Standard normal
distribution
0.0013 0.0214
0.1360
0.1360
0.0214 0.0013
–3 –2 –1 0 1 2 3 Z
(c)
Several probability statements for Z and X are summarized in the following table and are
shown on the distribution curve for Z in Figure 19–10 c .
Variable X Range Probability Variable Z Range
1 0.3413 0 to 1
1 0.6826 1 to 1
2 0.4773 0 to 2
2 0.9546 2 to 2
3 0.4987 0 to 3
3 0.9974 3 to 3
As an illustration, probability statements from this tabulation and Figure 19–10 c for X and Z
are as follows:
The probability that X is within 2 of its mean is 0.9546.
The probability that Z is within 2 of its mean, which is the same as between the values
2 and 2, is also 0.9546.

19.5 Monte Carlo Sampling and Simulation Analysis 533
In order to take a random sample from a normal N ( µ , 2 ) population, a specially prepared table of
SND random numbers is used. (Tables of SND values are available in many statistics books.) The
numbers are actually values from the Z or N (0, 1) distribution and have values such as 2.10,
1.24, etc. Translation from the Z value back to the sample values for X is via Equation [19.19].
Interpretation of the home office memo
The statement that virtually all the local contract amounts should be between $8000 and
$12,000 may be interpreted as follows: A normal distribution is assumed with a mean of µ
$10,000 and a standard deviation for $667, or a variance of 2 ($667) 2 ; that is, an
N [$10,000, ($667)
2
] distribution is assumed. The value $667 is calculated using the fact
that virtually all the probability (99.74%) is within 3 of the mean, as stated above. Therefore,
3 $2000 and $667 (rounded off)
As an illustration, if six SND random numbers are selected and used to take a sample of size 6
from the normal distribution N [$10,000, ($667)
2
], the results are as follows:
SND Random Number
Z
X Using Equation [19.19]
X Z
2.10 X (2.10)(667) 10,000 $8599
3.12 X (3.12)(667) 10,000 $12,081
0.23 X (0.23)(667) 10,000 $9847
1.24 X (1.24)(667) 10,000 $10,827
2.61 X (2.61)(667) 10,000 $8259
0.99 X (0.99)(667) 10,000 $9340
In this sample of six typical concrete surfacing contract amounts for sites in our region, the
average is $9825 and five of six values are within the range of $8000 to $12,000, with the sixth
being only $81 above the upper limit.
19.5 Monte Carlo Sampling and Simulation Analysis
Up to this point, all alternative selections have been made using estimates with certainty, possibly
followed by some testing of the decision via sensitivity analysis or expected values. In this
section, we will use a simulation approach that incorporates the material of the previous sections
to facilitate the engineering economy decision about one alternative or between two or
more alternatives.
The random sampling technique discussed in Section 19.3 is called Monte Carlo sampling .
The general procedure outlined below uses Monte Carlo sampling to obtain samples of size n for
selected parameters of formulated alternatives. These parameters, expected to vary according to a
stated probability distribution, warrant decision making under risk. All other parameters in an alternative
are considered certain; that is, they are known, or they can be estimated with enough precision
to consider them certain. An important assumption is made, usually without realizing it.
All parameters are independent; that is, one variable’s distribution does not affect the value of
any other variable of the alternative. This is referred to as the property of independent random
variables.
The simulation approach to engineering economy analysis is summarized in the following
basic steps.
Step 1. Formulate alternative(s). Set up each alternative in the form to be considered
using engineering economic analysis, and select the measure of worth upon which
to base the decision. Determine the form of the relation(s) to calculate the measure
of worth.
Step 2. Parameters with variation. Select the parameters in each alternative to be treated
as random variables. Estimate values for all other (certain) parameters for the analysis.

534 Chapter 19 More on Variation and Decision Making under Risk
Step 3. Determine probability distributions. Determine whether each variable is discrete
or continuous, and describe a probability distribution for each variable in each
alternative. Use standard distributions, where possible, to simplify the sampling process
and to prepare for computer-based simulation.
Step 4. Random sampling. Incorporate the random sampling procedure of Section 19.3
(the first four steps) into this procedure. This results in the cumulative distribution,
assignment of RNs, selection of the RNs, and a sample of size n for each variable.
Step 5. Measure of worth calculation. Compute n values of the selected measure of
worth from the relation(s) determined in step 1. Use the estimates made with certainty
and the n sample values for the varying parameters. (This is when the property
of independent random variables is actually applied.)
Step 6. Measure of worth description. Construct the probability distribution of the measure
__ of __ worth, using between 10 and 20 cells of data, and calculate measures such as
X , s , X ts , and relevant probabilities.
Step 7. Conclusions. Draw conclusions about each alternative, and decide which is to be
selected. If the alternative(s) has (have) been previously evaluated under the assumption
of certainty for all parameters, comparison of results may help with the final decision.
Example 19.8 illustrates this procedure using an abbreviated manual simulation analysis, and
Example 19.9 utilizes spreadsheet simulation for the same estimates.
EXAMPLE 19.8
Yvonne Ramos is the CEO of a chain of 50 fitness centers in the United States and Canada. An
equipment salesperson has offered Yvonne two long-term opportunities on new aerobic exercise
systems, for which the usage is charged to customers on a per-use basis on top of the
monthly fees paid by customers. As an enticement, the offer includes a guarantee of annual
revenue for one of the systems for the first 5 years.
Since this is an entirely new and risky concept of revenue generation, Yvonne wants to do a
careful analysis of each alternative. Details for the two systems follow:
System 1. First cost is P $12,000 for a set period of n 7 years with no salvage value.
No guarantee for annual net revenue is offered.
System 2. First cost is P $8000, there is no salvage value, and there is a guaranteed
annual net revenue of $1000 for each of the first 5 years, but after this period, there is
no guarantee. The equipment with updates may be useful up to 15 years, but the exact
number is not known. Cancellation anytime after the initial 5 years is allowed, with no
penalty.
For either system, new versions of the equipment will be installed with no added costs. If the
MARR is 15% per year, use PW analysis to determine if neither, one, or both of the systems
should be installed.
Solution by Hand
Estimates that Yvonne makes to use the simulation analysis procedure are included in the following
steps.
Step 1. Formulate alternatives. Using PW analysis, the relations for system 1 and system
2 are developed. The symbol NCF identifies the net cash flows (revenues), and
NCF G is the guaranteed NCF of $1000 for system 2.
PW 1 P 1 NCF 1 ( P A ,15%, n 1 ) [19.20]
PW 2 P 2 NCF G ( P A ,15%,5)
NCF 2 ( P A ,15%, n 2 5)( P F ,15%,5) [19.21]
Step 2. Parameters with variation. Yvonne summarizes the parameters estimated with
certainty and makes distribution assumptions about three parameters treated as random
variables.

19.5 Monte Carlo Sampling and Simulation Analysis 535
System 1
Certainty. P 1 $12,000; n 1 7 years.
Variable. NCF 1 is a continuous variable, uniformly distributed between
L $−4000 and H $6000 per year, because this is considered a high-risk
venture.
System 2
Certainty. P 2 $8000; NCF G $1000 for first 5 years.
Variable. NCF 2 is a discrete variable, uniformly distributed over the
values L $1000 to H $6000 only in $1000 increments, that is, $1000,
$2000, etc.
Variable. n 2 is a continuous variable that is uniformly distributed between
L 6 and H 15 years.
Now, rewrite Equations [19.20] and [19.21] to reflect the estimates made with certainty.
PW 1 12,000 NCF 1 ( P A ,15%,7)
12,000 NCF 1 (4.1604) [19.22]
PW 2 8000 1000( P A ,15%,5)
NCF 2 ( P A ,15%, n 2 5)( P F ,15%,5)
4648 NCF 2 ( P A ,15%, n 2 5)(0.4972) [19.23]
Step 3. Determine probability distributions. Figure 19–11 (left side) shows the assumed
probability distributions for NCF 1 , NCF 2 , and n 2 .
Step 4. Random sampling. Yvonne decides on a sample of size 30 and applies the first
four of the random sample steps in Section 19.3. Figure 19–11 (right side) shows the
cumulative distributions (step 1) and assigns RNs to each variable (step 2). The RNs
for NCF 2 identify the x axis values so that all net cash flows will be in even $1000
amounts. For the continuous variable n 2 , three-digit RN values are used to make the
numbers come out evenly, and they are shown in cells only as “indexers” for easy
reference when a RN is used to find a variable value. However, we round the number
to the next higher value of n 2 because it is likely the contract may be canceled on an
anniversary date. Also, now the tabulated compound interest factors for ( n 2 5)
years can be used directly (see Table 19–5 ).
Once the first RN is selected randomly from Table 19–2 , the sequence (step 3)
used will be to proceed down the RN table column and then up the column to the left.
Table 19–5 shows only the first five RN values selected for each sample and the corresponding
variable values taken from the cumulative distributions in Figure 19–11
(step 4).
Step 5. Measure of worth calculation. With the five sample values in Table 19–5 , calculate
the PW values using Equations [19.22] and [19.23].
1. PW 1 12,000 (2200)(4.1604) $21,153
2. PW 1 12,000 2000(4.1604) $3679
3. PW 1 12,000 (1100)(4.1604) $16,576
4. PW 1 12,000 (900)(4.1604) $15,744
5. PW 1 12,000 3100(4.1604) $897
1. PW 2 4648 1000( P A ,15%,7)(0.4972) $2579
2. PW 2 4648 1000( P A ,15%,5)(0.4972) $2981
3. PW 2 4648 5000( P A ,15%,8)(0.4972) $6507
4. PW 2 4648 3000( P A ,15%,10)(0.4972) $2838
5. PW 2 4648 4000( P A ,15%,3)(0.4972) $107
Now, 25 more RNs are selected for each variable from Table 19–2 , and the PW values
are calculated.
Step 6. Measure of worth description. Figure 19–12 a and b presents the PW 1 and PW 2
probability distributions for the 30 samples with 14 and 15 cells, respectively, as well
as the range of individual PW values and the X __
and s values.

536 Chapter 19 More on Variation and Decision Making under Risk
Figure 19-11
Distributions used for
random samples,
Example 19.8.
f (NCF 1 )
1
10
P(NCF 2 )
1
6
Continuous variable
–4 –2 0 2 4 6
NCF 1 , $1000
Discrete variable
F(NCF 1 )
1.0
0.8
0.6
0.4
0.2
0
F(NCF 2 )
1.00
–4 –2 0 2 4 6
0
0 1 2 3 4 5 6
1 2 3 4 5 6
NCF 2 , $1000 NCF 2 , $1000
0.83
0.67
0.50
0.33
0.17
NCF 1 , $1000
RN
80–99
60–79
40–59
20–39
00–19
RN
83–99
67–82
50–66
33–49
17–32
00–16
F(n 2 )
1.0
RN
888–999
0.8
666–887
0.6
444–665
1
9
f (n 2 )
Continuous variable
0.4
0.2
222–443
000–221
6 10 15
n 2 , years
0
6
8 10 12 14 15
n 2 , years
TABLE 19–5 Random Numbers and Variable Values for NCF 1 , NCF 2 , and
n 2 , Example 19.8
NCF 1 NCF 2 n 2
RN* Value, $ RN † Value, $ RN ‡ Value, Year Rounded §
18 2200 10 1000 586 11.3 12
59 2000 10 1000 379 9.4 10
31 1100 77 5000 740 12.7 13
29 900 42 3000 967 14.4 15
71 3100 55 4000 144 7.3 8
*Randomly start with row 1, column 4 in Table 19–2.
† Start with row 6, column 14.
‡ Start with row 4, column 6.
§ The n 2 value is rounded up.

19.5 Monte Carlo Sampling and Simulation Analysis 537
X 1 ± 1s 1
n 1 = 30
s 1 = $10,190
Frequency
X 1 = $–7729
Range = $37,443
5
4
PW 0
3
2
1
0
–27 –24 –21 –18 –15 –12 –9 –6 –3 0 3 6 9 12 15
PW 1 , $1000
(a) System 1
Frequency
6
5
4
X 2 ± 1s 2
X 2 = $2724
PW 0
n 2 = 30
s 2 = $4336
Range = $13,355
3
2
1
0
–4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10 11
PW 2 , $1000
(b) System 2
Figure 19–12
Probability distributions of simulated PW values for a sample of size 30, Example 19.8.
PW 1 . Sample values range from $−24,481 to $12,962. The calculated measures
of the 30 values are
__
X 1 $7729
s 1 $10,190
PW 2 . Sample values range from $−3031 to $10,324. The sample measures
are
__
X 2 $2724
s 2 $4336
Step 7. Conclusions. Additional sample values will surely make the central tendency of
the PW distributions more evident and may reduce the s values, which are quite
large. Of course, many conclusions are possible once the PW distributions are
known, but the following seem clear at this point.
System 1. Based on this small sample of 30 observations, do not accept this
alternative. The likelihood of making the MARR 15% is relatively small, since
the sample indicates a probability of 0.27 (8 out of 30 values) that the PW will be
positive, and X __
1 is a large negative. Though appearing large, the standard deviation
may be used to determine that about 20 of the 30 sample PW values (twothirds)
are within the limits X __
1 s , which are $17,919 and $2461. A larger
sample may alter this analysis somewhat.
System 2. If Yvonne is willing to accept the longer-term commitment that may
increase the NCF some years out, the sample of 30 observations indicates to accept
this alternative. At a MARR of 15%, the simulation approximates the chance
for a positive PW as 67% (20 of the 30 PW values in Figure 19–12 b are positive).
However, the probability of observing PW within the X __
1 s limits ($1612 and
$7060) is 0.53 (16 of 30 sample values).
Conclusion at this point. Reject system 1; accept system 2; and carefully
watch net cash flow, especially after the initial 5-year period.

538 Chapter 19 More on Variation and Decision Making under Risk
Comment
The estimates in Example 13.5 are very similar to those here, except all estimates were made
with certainty (NCF 1 $3000, NCF 2 $3000, and n 2 14 years). The alternatives were
evaluated by the payback period method at MARR 15%, and the first alternative was selected.
However, the subsequent PW analysis in Example 13.5 selected alternative 2 based, in
part, upon the anticipated larger cash flow in the later years.
EXAMPLE 19.9
Help Yvonne Ramos set up a spreadsheet simulation for the three random variables and PW
analysis in Example 19.8. Does the PW distribution vary appreciably from that developed
using manual simulation? Do the decisions to reject the system 1 proposal and accept the system
2 proposal still seem reasonable?
Solution by Spreadsheet
Figures 19–13 and 19–14 are spreadsheet screen shots that accomplish the simulation portion
of the analysis described above in steps 3 (determine probability distribution) through 6 (measure
of worth description). Most spreadsheet systems are limited in the variety of distributions
they can accept for sampling, but common ones such as uniform and normal are available.
Figure 19–13 shows the results of a small sample of 30 values from the three distributions
using the RAND and IF functions. (See Section A.3 in Appendix A.)
NCF 1 : Continuous uniform from $−4000 to $6000. The spreadsheet relation in column B
translates RN1 values (column A) into NCF1 amounts.
NCF 2 : Discrete uniform in $1000 increments from $1000 to $6000. Column D cells display
NCF2 in the $1000 increments using the logical IF function to translate from the RN2 values.
n 2 : Continuous uniform from 6 to 15 years. The results in column F are integer values obtained
using the INT function operating on the RN3 values.
Figure 19–13
Random sample of 30 values generated for spreadsheet simulation, Example 19.9.

19.5 Monte Carlo Sampling and Simulation Analysis 539
AVERAGE(F13:F42)
SUM(G13:G42)
STDEV(F13:F42)
Figure 19–14
Simulation results for 30 PW values, Example 19.9.
Figure 19–14 presents the two alternatives’ estimates in the top section. The PW1 and PW2
computations for the 30 repetitions of NCF1, NCF2, and N2 are the spreadsheet equivalent of
Equations [19.22] and [19.23]. The tabular approach used here tallies the number of PW values
below zero ($0) and equal to or exceeding zero using the IF operator. For example, cell C17
contains a 1, indicating PW1 0 when NCF1 $3100 (in cell B7 of Figure 19–13 ), which was
used to calculate PW1 $897 by Equation [19.22]. Cells in rows 7 and 8 show the number of
times in the 30 samples that system 1 and system 2 may return at least the MARR 15% because
the corresponding PW 0. Sample averages and standard deviations are also indicated.
Comparison between the hand and spreadsheet simulations is presented below.
System 1 PW
__
X , $ s , $
No. of
PW 0
System 2 PW
__
X , $ s , $
No. of
PW 0
Hand 7,729 10,190 8 2,724 4,336 20
Spreadsheet 7,105 13,199 10 1,649 3,871 19
For the spreadsheet simulation, 10 (33%) of the PW1 values exceed zero, while the manual simulation
included 8 (27%) positive values. These comparative results will change every time this
spreadsheet is activated since the RAND function is set up (in this case) to produce a new RN each
time. (It is possible to define RAND to keep the same RN values. See the Excel User’s Guide.)
The conclusion to reject the system 1 proposal and accept system 2 is still appropriate for
the spreadsheet simulation as it was for the hand solution, since there are comparable chances
that PW 0.

540 Chapter 19 More on Variation and Decision Making under Risk
CHAPTER SUMMARY
To perform decision making under risk implies that some parameters of an engineering alternative
are treated as random variables. Assumptions about the shape of the variable’s probability
distribution are used to explain how the estimates of parameter values may vary. Additionally,
measures such as the expected value and standard deviation describe the characteristic shape of
the distribution. In this chapter, we learned several of the simple, but useful, discrete and continuous
population distributions used in engineering economy—uniform and triangular—as well
as specifying our own distribution or assuming the normal distribution.
Since the population’s probability distribution for a parameter is not fully known, a random
sample of size n is usually taken, and its sample average and standard deviation are determined.
The results are used to make probability statements about the parameter, which help make the
final decision with risk considered.
The Monte Carlo sampling method is combined with engineering economy relations for a
measure of worth such as PW to implement a simulation approach to risk analysis. The results of
such an analysis can then be compared with decisions when parameter estimates are made with
certainty.
PROBLEMS
Certainty, Risk, and Uncertainty
19.1 Identify the following variables as either discrete
or continuous.
(a) The interest rates available in the marketplace
for jumbo certificates of deposit
(b)
Optimistic, most likely, and pessimistic estimates
of salvage value
(c) The number of cars that are red in the first
100 that pass through a certain intersection
(d ) The weight of the purse or wallet carried when
a person leaves her or his residence
(e) The gallons of water that evaporate from
Lake Erie in a given day
19.2 For each situation below, determine (1) if the variable
is discrete or continuous and (2) if the information
involves certainty, risk, andor uncertainty.
(a)
The first cost of a new front-end loader is
$34,000 or $38,000 depending on the size
purchased.
(b) The raises for engineers and technical staff
employees will be 3%, or 5%, with one-half
getting 3% and one-half getting 5%.
(c)
Revenue from a new product line is expected
to be between $350,000 and $475,000 per
year.
(d ) The salvage value for an old machine will be
$500 (i.e., its asking price) or $0 (it will be
thrown away).
(e)
Profits are equally likely to be up anywhere
from 25% to 60% this year.
19.3 An engineer learned that production output is between
1000 and 2000 units per week 90% of the
time, and it may fall below 1000 or go above
2000. He wants to use E (output) in the decisionmaking
process. Identify at least two additional
pieces of information that must be obtained or
assumed to finalize the output information for
this use.
Probability and Distributions
19.4 Royalties received by an investor in an oil well
vary according to the price of oil. Data collected
from stripper wells in an established oil field were
used to develop the probability-royalty relationship
shown below.
Royalties, $ per Year 6200 8500 9600 10,300 12,600 15,500
Probability 0.10 0.21 0.32 0.24 0.09 0.04
(a) Calculate the expected value of royalty income
(RI) per year.
(b) Determine the probability that the royalty
income will be at least $12,600 per year.
19.5 Daily revenue from vending machines placed in
various buildings of a major university is as follows:
20, 75, 43, 62, 51, 52, 78, 33, 28, 39, 61, 56, 43, 49, 48, 49,
71, 53, 57, 46, 42, 41, 63, 36, 51, 59, 40, 32, 37, 29, 26
(a) Construct a frequency distribution table with
a cell size of 12 starting with 19.5 (i.e., first
cell is 19.5–31.5, next is 31.5–43.5, etc.).
(b) Determine the probability distribution.
(c) What is the probability of revenue from a
machine being less than $44?
(d) What is the probability that revenue from a
machine will equal or exceed $44?

Problems 541
19.6 A survey of households included a question about
the number of operating automobiles N currently
owned by people living at the residence and the
interest rate i on the lowest-rate loan for the cars.
The results for 100 households are shown.
Number of Cars N
Households
0 12
1 56
2 26
3 3
4 3
Loan Rate i , % Households
0.02 13
2.014 14
4.016 19
6.018 38
8.0110 12
10.0112 4
( a) State whether each variable is discrete or
continuous.
( b) Plot the probability distributions and cumulative
distributions for N and i .
( c) From the data collected, what is the probability
that a household has 1 or 2 cars? Three or
more cars?
( d) Use the data for i to estimate the chances that
the interest rate is between 7% and 11% per
year.
19.7 An officer of the state lottery commission sampled
lottery ticket purchasers over a 1-week period at
one location. The amounts distributed back to the
purchasers and the associated probabilities for
5000 tickets are as follows:
Distribution, $ 0 2 5 10 100
Probability 0.91 0.045 0.025 0.013 0.007
( a) Plot the cumulative distribution of winnings.
( b) Calculate the expected value of the distribution
of dollars per ticket.
( c) If tickets cost $2, what is the expected longterm
income to the state per ticket, based
upon this sample?
19.8 Bob is working on two separate probability-related
projects. The first involves a variable N , which is the
number of consecutively manufactured parts that
weigh in above the weight specification limit. The
variable N is described by the formula (0.5) N because
each unit has a 50-50 chance of being below
or above the limit. The second involves a battery
life L that varies between 2 and 5 months. The probability
distribution is triangular with the mode at
5 months, which is the design life. Some batteries
fail early, but 2 months is the smallest life experienced
thus far. ( a ) Write out and plot the probability
distributions and cumulative distributions for Bob.
( b ) Determine the probability of N being 1, 2, or 3
consecutive units above the weight limit.
19.9 An alternative to buy and an alternative to lease
hydraulic lifting equipment have been formulated.
Use the parameter estimates and assumed distribution
data shown to plot the probability distributions
on one graph for each corresponding parameter.
Label the parameters carefully.
Parameter
Purchase Alternative
Estimated Value
High
Low
Assumed
Distribution
First cost, $ 25,000 20,000 Uniform;
continuous
Salvage value, $ 3,000 2,000 Triangular;
mode at $2500
Life, years 8 4 Triangular;
mode at 6
AOC, $ per year 9,000 5,000 Uniform;
continuous
Parameter
Lease Alternative
Estimated Value
High
Low
Assumed
Distribution
Lease first cost, $ 2000 1800 Uniform;
continuous
AOC, $ per year 9000 5000 Triangular;
mode at $7000
Lease term, years 3 3 Certainty
19.10 Carla is a statistician with a bank. She has collected
debt-to-equity mix data on mature ( M ) and
young ( Y ) companies. The debt percentages vary
from 20% to 80% in her sample. Carla has defined
D M as a variable for the mature companies from
0 to 1, with D M 0 interpreted as the low of 20%
debt and D M 1.0 as the high of 80% debt. The
variable for young corporation debt percentages
D Y is similarly defined. The probability distributions
used to describe D M and D Y are
f ( D M ) 3(1 D M ) 2 0 D M 1
f ( D Y ) 2 D Y 0 D Y 1
(a) Use different values of the debt percentage between
20% and 80% to calculate values for the probability
distributions and then plot them. ( b ) What can
you comment about the probability that a mature
company or a young company will have a low debt
percentage? A high debt percentage?

542 Chapter 19 More on Variation and Decision Making under Risk
19.11 A discrete variable X can take on integer values of
1 to 10. A sample of size 50 results in the following
probability estimates:
X i 1 2 3 6 9 10
P(X i ) 0.2 0.2 0.2 0.1 0.2 0.1
(a) Write out and graph the cumulative
distribution.
(b) Calculate the following probabilities using
the cumulative distribution: X is between 6
and 10, and X has the values 4, 5, or 6.
(c) Use the cumulative distribution to show that
P ( X 7 or 8) 0.0. Even though this probability
is zero, the statement about X is that it
can take on integer values of 1 to 10. How do
you explain the apparent contradiction in
these two statements?
Random Samples
19.12 A discrete variable X can take on integer values of
1 to 5. A sample of size 100 results in the following
probability estimates.
X i 1 2 3 4 5
P(X i ) 0.2 0.3 0.1 03 0.1
(a) Use the following random numbers to estimate
the probabilities for each value of X .
(b) Determine the sample probabilities for X 1
and X 5. Compare the sample results with
the probabilities in the problem statement.
RN: 10, 42, 18, 31, 23, 80, 80, 26, 74, 71, 03, 90,
55, 61, 61, 28, 41, 49, 00, 79, 96, 78, 42, 31, 26
19.13 The percent price increase p on a variety of retail
food prices over a 1-year period varied from 5% to
10% in all cases. Because of the distribution of
p values, the assumed probability distribution for
the next year is
where
f ( X ) 2 X 0 X 1
0 when p 5%
X
{ 1 when p 10%
For a continuous variable the cumulative distribution
F ( X ) is the integral of f ( X ) over the same
range of the variable. In this case
F ( X ) X 2 0 X 1
(a) Graphically assign RNs to the cumulative
distribution, and take a sample of size 30 for
the variable. Transform the X values into interest
rates.
(b) Calculate the average p value for the sample.
19.14 Develop a discrete probability distribution of your
own for the variable G , the expected grade in this
course, where G A, B, C, D, F, or I (incomplete).
Assign random numbers to F ( G ), and take a sample
from it. Now plot the probability values from
the sample for each G value.
19.15 Use the RAND function in Excel to generate
100 values from a U (0,1) distribution.
(a) Calculate the average and compare it to 0.5,
the expected value for a random sample between
0 and 1.
(b) For the RAND function sample, cluster the
results into cells of 0.1 width, that is 0.0–0.1,
0.1–0.2, etc., where the upper-limit value is
excluded from each cell. Determine the probability
for each grouping from the results.
Does your sample come close to having
approximately 10% in each cell?
Sample Estimates
19.16 An engineer was asked to determine whether the
average air quality in a vehicle assembly plant was
within OSHA guidelines. The following air quality
readings were collected:
(a)
(b)
(c)
81, 86, 80, 91, 83, 83, 96, 85, 89
Determine the sample mean.
Calculate the standard deviation.
Determine the number of values and percent
of values that fall within 1 standard deviation
of the mean.
19.17 Carol sampled the monthly maintenance costs for automated
soldering machines a total of 100 times during
1 year. She clustered the costs into $200 cells, for
example, $500 to $700, with cell midpoints of $600,
$800, $1000, etc. She indicated the number of times
(frequency) each cell value was observed. The costs
and frequency data are as follows.
Cell
Midpoint, $
Frequency
600 6
800 10
1000 7
1200 15
1400 28
1600 15
1800 9
2000 10
(a) Estimate the expected value and standard deviation
of the maintenance costs the company
should anticipate based on Carol’s sample.
( b) What is the best estimate of the percentage of
costs that will fall within 2 standard deviations
of the mean?

Additional Problems and FE Exam Review Questions 543
( c) Develop a probability distribution of the
monthly maintenance costs from Carol’s
sample, and indicate the answers to the previous
two questions on it.
(d) Use a spreadsheet to display the sample mean.
19.18 ( a ) Determine the values of sample average and
standard deviation of the data in Problem 19.11.
( b ) Determine the values 1 and 2 standard deviations
from the mean. Of the 50 sample points, how
many fall within these two ranges?
19.19 ( a ) Use the relations in Section 19.4 for continuous
variables to determine the expected value and
standard deviation for the distribution of f ( D Y ) in
Problem 19.10. ( b ) It is possible to calculate the
probability of a continuous variable X between
two points ( a, b ) using the following integral:
b
P ( a X b ) a f ( X ) dx
Determine the probability that D Y is within 2 standard
deviations of the expected value.
19.20 ( a ) Use the relations in Section 19.4 for continuous
variables to determine the expected value
and variance for the distribution of D M in
Problem 19.10.
f ( D M ) 3(1 D M ) 2 0 D M 1
(b) Determine the probability that D M is within 2
standard deviations of the expected value.
Use the relation in Problem 19.19.
19.21 Calculate the expected value for the variable N in
Problem 19.8.
19.22 A newsstand manager is tracking Y , the number of
weekly magazines left on the shelf when the new
edition is delivered. Data collected over a 30-week
period are summarized by the following probability
distribution. Plot the distribution and the estimates
for expected value and 1 standard deviation
on either side of E ( Y ) on the plot.
Simulation
19.23 Carl, an engineering colleague, estimated net cash
flow after taxes (CFAT) for the project he is working
on. The additional CFAT of $2800 in year 10 is
the salvage value of capital assets.
Year CFAT, $
0 28,800
16 5,400
710 2,040
10 2,800
The PW value at the current MARR of 7% per
year is
PW 28,800 5400( P A ,7%,6)
2040( P A ,7%,4)( P F ,7%,6)
2800( P F ,7%,10)
$2966
Carl believes the MARR will vary over a relatively
narrow range, as will the CFAT, especially during
the out years of 7 through 10. He is willing to accept
the other estimates as certain. Use the following
probability distribution assumptions for MARR
and CFAT to perform a simulation—hand- or
spreadsheet-based.
MARR. Uniform distribution over the range
6% to 10%.
CFAT, years 7 through 10. Uniform distribution
over the range $1600 to $2400 for each
year.
Plot the resulting PW distribution. Should the plan
be accepted using decision making under certainty?
Under risk?
19.24 Repeat Problem 19.23, except use the normal distribution
for the CFAT in years 7 through 10 with
an expected value of $2000 and a standard deviation
of $500.
Y copies 3 7 10 12
P ( Y ) 13 14 13 112
ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS
19.25 When there are at least two values for a parameter
and it is possible to estimate the chance that each
may occur, this situation is known as:
( a) Uncertainty
( b) Risk
( c) Standard deviation
( d) Cost estimating
19.26 A deterministic economic analysis is one wherein:
(a) Single-value estimates are used exclusively
(b) Risk is taken into account
(c) A range of values for each parameter is included
in the analysis
(d) Taxes and inflation are not considered in the
cash flow estimates

544 Chapter 19 More on Variation and Decision Making under Risk
19.27 Decision making under risk includes all of the following
except:
(a) Expected value analysis
(b) Simulation
(c) Using only single-value estimates
(d) Probability
19.28 All of the following are elements in decision making
under risk except:
(a) Random variable
(b) Cost indexes
(c) Probability
(d) Cumulative distribution
19.29 For the income and probability values shown in
the table, the probability that the income in any
year will be less than $9600 is closest to:
(a) 0.15
(b) 0.23
(c) 0.32
(d) 0.38
Income, $ per Year 6200 8500 9600 10,300 12,600 15,500
Probability 0.15 0.23 0.32 0.24 0.09 0.04
19.30 The symbol that represents the true population
mean is:
(a)
(b) s
(c) __
(d) X
19.31 The revenue from an oil dispersant product has
averaged $15,000 per month for the past 12 months.
If the value of ( X i − X __
) 2 is $1,600,000, the standard
deviation is closest to:
( a) $381
( b) $652
( c) $958
( d) $1265
19.32 A survey of the types of cars parked at an NFL
football stadium revealed that there were equal
probabilities of finding cars identified as type A, B,
C, and D. Car types were assigned random numbers
as follows.
Car Type
Assigned RN
A 0 through 24
B 25 through 49
C 50 through 74
D 75 through 99
The sample probability of a type B car from the
12 random numbers shown is closest to:
RN: 75, 52, 14, 99, 67, 74, 06, 50, 97, 46, 27, 88
(a) 0.17
(b) 0.25
(c) 0.33
(d ) 0.42
CASE STUDY
USING SIMULATION AND THREE-ESTIMATE SENSITIVITY ANALYSIS
Background
The Knox Brewing company makes specialty-named sodas
and flavored drinks for retail grocery chains throughout
North and Central America. During the past year, it has become
obvious that a new bottle-capping machine is needed to
replace the current 10-year-old system. Dr. Knox, the owner
and president, knows the business quite well. You just handed
him the first cost bids from three vendors for the machine. He
looked carefully at the numbers and asked you to sit down.
You were quite surprised, as this was the first time you had
been in his office, and most other engineers at Knox have a
great fear of “the Old Man.”
Information
As he examined the three bids on first cost of the machine, he
started to write some numbers, which, he explained, were his
estimates of the annual operating cost, useful life, and possible
salvage value for each of the machines sold by the three vendors.
After a few minutes, he told you to take these numbers
and use some of that “new engineering knowledge” you
acquired in college to determine which, if any, of these three
bids made the best economic sense. He also told you to be
innovative and use a computer and some probability to come
up with a robust recommendation by tomorrow at 2 P.M.
You have used the estimates from the president to develop
Table 19–6 of pessimistic (P), most likely (ML), and optimistic
(O) estimates for each vendor’s machine. In addition, you
developed some possible distributions for the parameters that
Dr. Knox estimated, namely, AOC, life, and salvage value.
These are summarized in Table 19–7 . You plan to use a simple
Monte Carlo simulation to help formulate your recommendation
for tomorrow.
Case Study Exercises
First, learn to use the RNG (random number generator) in
Excel, if you have not already done so. It is necessary to

Case Study 545
TABLE 19–6 Parameter Estimates for Bottle-Capping Machine
First Cost, $ AOC, $ per Year Salvage, $ Life, Years
Vendor 1
P 200,000 11,000 0 3
ML 200,000 10,000 0 5
O −200,000 6,000 0 8
Vendor 2
P 150,000 5,000 0 2
ML 150,000 3,500 5,000 4
O 150,000 2,000 8,000 7
Vendor 3
P 300,000 8,000 5,000 5
ML 300,000 6,000 5,000 7
O 300,000 4,500 8,000 9
TABLE 19–7 Distribution Assumptions about AOC, Life, and Salvage
Parameter Vendor 1 Vendor 2 Vendor 3
AOC, $ per year
Salvage, $
Life, years
Normal
Mean: 10,000
Std. dev.: 500
Uniform
0 to 1000
Discrete uniform
3 to 8, equal probability
Normal
Mean: 3500
Std. dev.: 500
Uniform
0 to 8000
Discrete uniform
2 to 7, equal probability
Normal
Mean: 6000
Std. dev.: 500
Uniform
5000 to 8000
Discrete uniform
5 to 9, equal probability
sample from the normal distributions that you have specified
in Table 19–7 . RNG is part of the Analysis Tool-Pak
accessed through the Office Button, Excel Options, Add-
Ins path.
1. Prepare the simulation using a spreadsheet; determine
which of the vendors offers the best machine from an
economic perspective, and take into account the estimates
made by Dr. Knox. Use a sample size of at least
50, and base your conclusions on the AW measure of
worth.
2. Prepare a short presentation for Dr. Knox (and class)
using your analysis.

APPENDIX A
USING SPREADSHEETS
AND MICROSOFT EXCEL ©
This appendix explains the layout of a spreadsheet and the use of Microsoft Excel (hereafter
called Excel) functions in engineering economy. Refer to the Excel help system for your particular
computer and version of Excel. Some specific commands and entries refer to Excel 2007 and
may differ slightly from your version.
A.1 Introduction to Using Excel
Enter a Formula or Use an Excel Function
The sign is required to perform any formula or function computation in a cell. The formulas
and functions on the worksheet can be displayed by simultaneously pressing Ctrl and `. The symbol
` is usually in the upper left of the keyboard with the ~ (tilde) symbol. Pressing Ctrl+` a second
time hides the formulas and functions.
1. Run Excel.
2. Move to cell C3. (Move the pointer to C3 and left-click.)
3. Type PV(5%,12,10) and Enter. This function will calculate the present value of
12 payments of $10 at a 5% per year interest rate.
Another example: To calculate the future value of 12 payments of $10 at 6% per year interest, do
the following:
1. Move to cell B3, and type INTEREST.
2. Move to cell C3, and type 6% or 6100.
3. Move to cell B4, and type PAYMENT.
4. Move to cell C4, and type 10 (to represent the size of each payment).
5. Move to cell B5, and type NUMBER OF PAYMENTS.
6. Move to cell C5, and type 12 (to represent the number of payments).
7. Move to cell B7, and type FUTURE VALUE.
8. Move to cell C7, and type FV(C3,C5,C4) and hit Enter. The answer will appear
in cell C7.
To edit the values in cells
1. Move to cell C3 and type 5100 (the previous value will be replaced).
2. The value in cell C7 will update.
Cell References in Formulas and Functions
If a cell reference is used in lieu of a specific number, it is possible to change the number once
and perform sensitivity analysis on any variable that is referenced by the cell number, such as C5.
This approach defines the referenced cell as a global variable for the worksheet. There are two
types of cell references—relative and absolute.
Relative References If a cell reference is entered, for example, A1, into a formula or function
that is copied or dragged into another cell, the reference is changed relative to the movement of
the original cell. If the formula in C5 is A1 and it is copied into cell C6, the formula is changed
to A2. This feature is used when dragging a function through several cells, and the source entries
must change with the column or row.
Absolute References If adjusting cell references is not desired, place a $ sign in front of
the part of the cell reference that is not to be adjusted—the column, row, or both. For example,

548 Appendix A Using Spreadsheets and Microsoft Excel ©
$A$1 will retain the formula when it is moved anywhere on the worksheet. Similarly, $A1
will retain the column A, but the relative reference on 1 will adjust the row number upon movement
around the worksheet.
Absolute references are used in engineering economy for sensitivity analysis of parameters
such as MARR, first cost, and annual cash flows. In these cases, a change in the absolutereference
cell entry can help determine the sensitivity of a result, such as PW or AW.
Print the Spreadsheet
First define the portion (or all) of the spreadsheet to be printed.
1. Move the pointer to the top left cell of your spreadsheet.
2. Hold down the left-click button. (Do not release the left-click button.)
3. Drag the mouse to the lower right corner of your spreadsheet or to wherever you want to stop
printing.
4. Release the left-click button. (It is ready to print.)
5. Left-click the Office button (see Figure A–1 ).
6. Move the pointer down to select Print and left-click.
7. In the dialog box, left-click the Print option (or similar command).
Depending on your computer environment, you may have to select a network printer and queue
your printout through a server.
Save the Spreadsheet
You can save your spreadsheet at any time during or after completing your work. It is recommended
that you save your work regularly.
1. Left-click the Office button.
2. To save the spreadsheet the first time, left-click the Save As . . . option.
3. Type the file name, e.g., Prob 7.9, and left-click the Save button.
To save the spreadsheet after it has been saved the first time, i.e., a file name has been assigned
to it, left-click the Office button, move the pointer down, and left-click on Save.
Create an xy (Scatter) Chart
This chart is one of the most commonly used in scientific analysis, including engineering economy.
It plots pairs of data and can place multiple series of entries on the Y axis. The xy scatter
chart is especially useful for results such as the PW versus i graph, where i is the X axis and the
Y axis displays the results of the NPV function for several alternatives.
1. Run Excel.
2. Enter the following numbers in columns A, B, and C, respectively.
Column A, cell A1 through A6: Rate i %, 4, 6, 8, 9, 10
Column B, cell B1 through B6: $ for A, 40, 55, 60, 45, 10
Column C, cell C1 through C6: $ for B, 100, 70, 65, 50, 30.
3. Move the mouse to A1, left-click, and hold while dragging to cell C6. All cells will be highlighted,
including the title cell for each column.
4. If not all the columns for the chart are adjacent to one another, first press and hold the Control
key on the keyboard during the entirety of step 3. After dragging over one column of
data, momentarily release the left click, then move to the top of the next (nonadjacent) column
of the chart. Do not release the Control key until all columns to be plotted have been
highlighted.
5. Left-click on the Insert button on the toolbar.
6. Select the Scatter option and choose a subtype of scatter chart. The graph appears with a
legend ( Figure A–1 ).
Now a large number of styling effects can be introduced for axis titles, legend, data series, etc.
Note that only the bottom row of the title can be highlighted. If titles are not highlighted, the data
sets are generically identified as series 1, series 2, etc. on the legend.

A.2 Organization (Layout) of the Spreadsheet 549
Office button
Insert for
scatter chart
Figure A–1
Scatter chart for data entries and location of commonly used buttons.
Obtain Help While Using Excel
1. To get general help information, left-click on the “?” (upper right).
2. Enter the topic or phrase. For example, if you want to know more about how to save a file,
type the word Save.
3. Select the appropriate matching words. You can browse through the options by left-clicking
on any item.
A.2 Organization (Layout) of the Spreadsheet
A spreadsheet can be used in several ways to obtain answers to numerical questions. The first is
as a rapid solution tool, often with the entry of only a few numbers or one predefined function.
For example, to find the future worth in a single-cell operation, move the pointer to any cell and
enter FV(8%,5,-2500). The display of $14,666.50 is the 8% future worth at the end of year 5
of five equal payments of $2500 each.
A second use is more formal; it presents data, solutions, graphs, and tables developed on the
spreadsheet and ready for presentation to others. Some fundamental guidelines in spreadsheet
organization are presented here. A sample layout is presented in Figure A–2 . As solutions become
more complex, organization of the spreadsheet becomes increasingly important, especially for
presentation to an audience via PowerPoint or similar software.
Cluster the data and the answers. It is advisable to organize the given or estimated data in the
top left of the spreadsheet. A very brief label should be used to identify the data, for example,
MARR in cell A1 and the value, 12%, in cell B1. Then B1 can be the referenced cell for all
entries requiring the MARR. Additionally, it may be worthwhile to cluster the answers into one
area and frame it. Often, the answers are best placed at the bottom or top of the column of entries
used in the formula or predefined function.
Enter titles for columns and rows. Each column or row should be labeled so its entries are
clear to the reader. It is very easy to select from the wrong column or row when no brief title is
present at the head of the data.
Enter income and cost cash fl ows separately. When there are both income and cost cash flows
involved, it is strongly recommended that the cash flow estimates for revenue (usually positive)

550 Appendix A Using Spreadsheets and Microsoft Excel ©
Function used to find FW
Figure A–2
Spreadsheet layout with cash flow estimates, results of functions, function formula detailed, and a scatter chart.
and first cost, salvage value, and annual costs (usually negative, with salvage a positive number)
be entered into two adjacent columns. Then a formula combining them in a third column displays
the net cash flow. There are two immediate advantages to this practice: fewer errors are made
when performing the summation and subtraction mentally, and changes for sensitivity analysis
are more easily made.
Use cell references. The use of absolute and relative cell references is a must when any changes
in entries are expected. For example, suppose the MARR is entered in cell B1 and three separate
references are made to the MARR in functions on the spreadsheet. The absolute cell reference
entry $B$1 in the three functions allows the MARR to be changed one time, not three.
Obtain a fi nal answer through summing and embedding. When the formulas and functions
are kept relatively simple, the final answer can be obtained using the SUM function. For example,
if the present worth values (PW) of two columns of cash flows are determined separately,
then the total PW is the SUM of the subtotals. This practice is especially useful when the cash
flow series are complex.
Prepare for a chart. If a chart (graph) will be developed, plan ahead by leaving sufficient
room on the right of the data and answers. Charts can be placed on the same worksheet or on a
separate worksheet. Placement on the same worksheet is recommended, especially when the
results of sensitivity analysis are plotted.
A.3 Excel Functions Important to Engineering
Economy (alphabetical order)
DB (Declining Balance)
Calculates the depreciation amount for an asset for a specified period n using the declining balance
method. The depreciation rate d used in the computation is determined from asset values
S (salvage value) and B (basis or first cost) as d 1 ( S B )
1 n
. This is Equation [16.12]. Threedecimal-place
accuracy is used for d .
cost
salvage
life
period
month
DB(cost, salvage, life, period, month)
First cost or basis of the asset.
Salvage value.
Depreciation life (recovery period).
The period, year, for which the depreciation is to be calculated.
(optional entry) If this entry is omitted, a full year is assumed for the
first year.

A.3 Excel Functions Important to Engineering Economy 551
Example A new machine costs $100,000 and is expected to last 10 years. At the end of
10 years, the salvage value of the machine is $50,000. What is the depreciation of the machine in
the first year and the fifth year?
Depreciation for the first year: DB(100000,50000,10,1)
Depreciation for the fifth year: DB(100000,50000,10,5)
Because of the manner in which the DB function determines the fixed percentage d and the accuracy
of the computations, it is recommended that the DDB function (below) be used for all declining
balance depreciation rates. Simply use the optional factor entry for rates other than d 2 n .
DDB (Double Declining Balance)
Calculates the depreciation of an asset for a specified period n using the double declining balance
method. A factor can also be entered for some other declining balance depreciation method by
specifying a factor in the function.
cost
salvage
life
period
factor
DDB(cost, salvage, life, period, factor)
First cost or basis of the asset.
Salvage value of the asset.
Depreciation life.
The period, a year, for which the depreciation is to be calculated.
(optional entry) If this entry is omitted, the function will use a double
declining method with 2 times the straight line rate. If, for example,
the entry is 1.5, the 150% declining balance method will be used.
Example A new machine costs $200,000 and is expected to last 10 years. The salvage value is
$10,000. Calculate the depreciation of the machine for the first and the eighth years. Finally,
calculate the depreciation for the fifth year using the 175% declining balance method.
Depreciation for the first year: DDB(200000,10000,10,1)
Depreciation for the eighth year: DDB(200000,10000,10,8)
Depreciation for the fifth year using 175% DB: DDB(200000,10000,10,5,1.75)
EFFECT (Effective Interest Rate)
Calculates the effective annual interest rate for a stated nominal annual rate and a given number
of compounding periods per year. Excel uses Equation [4.7] to calculate the effective rate.
EFFECT(nominal, npery)
nominal Nominal interest rate for the year .
npery Number of times interest is compounded per year.
Example Claude has applied for a $10,000 loan. The bank officer told him that the interest rate
is 8% per year and that interest is compounded monthly to conveniently match his monthly payments.
What effective annual rate will Claude pay?
Effective annual rate: EFFECT(8%,12)
EFFECT can also be used to find effective rates other than annually. Enter the nominal rate for
the time period of the required effective rate; npery is the number of times compounding occurs
during the time period of the effective rate.
Example Interest is stated as 3.5% per quarter with quarterly compounding. Find the effective
semiannual rate.
The 6-month nominal rate is 7%, and compounding is 2 times per 6 months.
Effective semiannual rate: EFFECT(7%,2)

552 Appendix A Using Spreadsheets and Microsoft Excel ©
FV (Future Value)
Calculates the future value (worth) based on periodic payments at a specific interest rate.
FV(rate, nper, pmt, pv, type)
rate
nper
pmt
pv
type
Interest rate per compounding period.
Number of compounding periods.
Constant payment amount.
The present value amount. If pv is not specified, the function will
assume it to be 0.
(optional entry) Either 0 or 1. A 0 represents payments made at the
end of the period, and 1 represents payments at the beginning of the
period. If omitted, 0 is assumed.
Example Jack wants to start a savings account that can be increased as desired. He will deposit
$12,000 to start the account and plans to add $500 to the account at the beginning of each month
for the next 24 months. The bank pays 0.25% per month. How much will be in Jack’s account at
the end of 24 months?
Future value in 24 months: FV(0.25%,24,500,12000,1)
IF (IF Logical Function)
Determines which of two entries is entered into a cell based on the outcome of a logical check on
the outcome of another cell. The logical test can be a function or a simple value check, but it must
use an equality or inequality sense. If the response is a text string, place it between quote marks
(“ ”). The responses can themselves be IF functions. Up to seven IF functions can be nested for
very complex logical tests.
IF(logical_test,value_if_true,value_if_false)
logical_test
value_if_true
value_if_false
Any worksheet function can be used here, including a
mathematical operation.
Result if the logical_test argument is true.
Result if the logical_test argument is false.
Example The entry in cell B4 should be “selected” if the PW value in cell B3 is greater than
or equal to zero and “rejected” if PW 0.
Entry in cell B4: IF(B3 0,“selected”,“rejected”)
Example The entry in cell C5 should be “selected” if the PW value in cell C4 is greater than
or equal to zero, “rejected” if PW 0, and “fantastic” if PW 200.
Entry in cell C5:
IF(C40,“rejected”,IF(C4200,“fantastic”,“selected”))
IPMT (Interest Payment)
Calculates the interest accrued for a given period n based on constant periodic payments and
interest rate.
IPMT(rate, per, nper, pv, fv, type)
rate Interest rate per compounding period.
per Period for which interest is to be calculated.
nper Number of compounding periods.
pv Present value. If pv is not specified, the function will assume it to be 0.
fv Future value. If fv is omitted, the function will assume it to be 0. The fv
can also be considered a cash balance after the last payment is made.
type (optional entry) Either 0 or 1. A 0 represents payments made at the
end of the period, and 1 represents payments made at the beginning of
the period. If omitted, 0 is assumed.

A.3 Excel Functions Important to Engineering Economy 553
Example Calculate the interest due in the 10th month for a 48-month, $20,000 loan. The interest
rate is 0.25% per month.
Interest due: IPMT(0.25%,10,48,20000)
IRR (Internal Rate of Return)
Calculates the internal rate of return between 100% and infinity for a series of cash flows at
regular periods.
IRR(values, guess)
values
guess
A set of numbers in a spreadsheet column (or row) for which the rate of
return will be calculated. The set of numbers must consist of at least one
positive and one negative number. Negative numbers denote a payment
made or cash outflow, and positive numbers denote income or cash
inflow.
(optional entry) To reduce the number of iterations, a guessed rate of
return can be entered. In most cases, a guess is not required, and a
10% rate of return is initially assumed. If the #NUM! error appears, try
using different values for guess. Inputting different guess values makes
it possible to determine the multiple roots for the rate of return equation
of a nonconventional cash flow series.
Example John wants to start a printing business. He will need $25,000 in capital and anticipates
that the business will generate the following incomes during the first 5 years. Calculate his
rate of return after 3 years and after 5 years.
Year 1 $5,000
Year 2 $7,500
Year 3 $8,000
Year 4 $10,000
Year 5 $15,000
Set up an array in the spreadsheet.
In cell A1, type 25000 (negative for payment).
In cell A2, type 5000 (positive for income).
In cell A3, type 7500.
In cell A4, type 8000.
In cell A5, type 10000.
In cell A6, type 15000.
Therefore, cells A1 through A6 contain the array of cash flows for the first 5 years, including the
capital outlay. Note that any years with a zero cash fl ow must have a zero entered to ensure that
the year value is correctly maintained for computation purposes.
To calculate the internal rate of return after 3 years, move to cell A7, and type IRR(A1:A4).
To calculate the internal rate of return after 5 years and specify a guess value of 5%, move to
cell A8, and type IRR(A1:A6,5%).
MIRR (Modified Internal Rate of Return)
Calculates the modified internal rate of return for a series of cash flows and reinvestment of income
and interest at a stated rate.
MIRR(values, finance_rate, reinvest_rate)
values
Refers to an array of cells in the spreadsheet. Negative
numbers represent payments, and positive numbers represent

554 Appendix A Using Spreadsheets and Microsoft Excel ©
fi nance_rate
reinvest_rate
income. The series of payments and income must occur at
regular periods and must contain at least one positive number
and one negative number.
Interest rate on funds borrowed from external sources
( i b in Equation [7.9]).
Interest rate for reinvestment on positive cash flows ( i i in
Equation [7.9]). (This is not the same reinvestment rate on the
net investments when the cash flow series is nonconventional.
See Section 7.5 for comments.)
Example Jane opened a hobby store 4 years ago. When she started the business, Jane borrowed
$50,000 from a bank at 12% per year. Since then, the business has yielded $10,000 the
first year, $15,000 the second year, $18,000 the third year, and $21,000 the fourth year. Jane
reinvests her profits, earning 8% per year. What is the modified rate of return after 3 years and
after 4 years?
In cell A1, type 50000.
In cell A2, type 10000.
In cell A3, type 15000.
In cell A4, type 18000.
In cell A5, type 21000.
To calculate the modified rate of return after 3 years, move to cell A6, and type
MIRR(A1:A4,12%,8%).
To calculate the modified rate of return after 4 years, move to cell A7, and type
MIRR(A1:A5,12%,8%).
NOMINAL (Nominal Interest Rate)
Calculates the nominal annual interest rate for a stated effective annual rate and a given number
of compounding periods per year. This function is designed to display only nominal annual rates.
NOMINAL(effective, npery)
effective Effective interest rate for the year .
npery Number of times that interest is compounded per year.
Example Last year, a corporate stock earned an effective return of 12.55% per year. Calculate
the nominal annual rate, if interest is compounded quarterly and compounded continuously.
Nominal annual rate, quarterly compounding: NOMINAL(12.55%,4)
Nominal annual rate, continuous compounding: NOMINAL(12.55%,100000)
NPER (Number of Periods)
Calculates the number of periods for the present worth of an investment to equal the future value
specified, based on uniform regular payments and a stated interest rate.
NPER(rate, pmt, pv, fv, type)
rate
pmt
pv
fv
type
Interest rate per compounding period.
Amount paid during each compounding period.
Present value (lump-sum amount).
(optional entry) Future value or cash balance after the last payment.
If fv is omitted, the function will assume a value of 0.
(optional entry) Enter 0 if payments are due at the end of the
compounding period and 1 if payments are due at the beginning of the
period. If omitted, 0 is assumed.

A.3 Excel Functions Important to Engineering Economy 555
Example Sally plans to open a savings account that pays 0.25% per month. Her initial deposit
is $3000, and she plans to deposit $250 at the beginning of every month. How many payments
does she have to make to accumulate $25,000 to buy a new car?
Number of payments: NPER(0.25%,250,3000,25000,1)
NPV (Net Present Value)
Calculates the net present value of a series of future cash flows at a stated interest rate.
NPV(rate, series)
rate
series
Interest rate per compounding period.
Series of costs and incomes set up in a range of cells in the
spreadsheet.
Example Mark is considering buying a sports store for $100,000 and expects to receive the
following income during the next 6 years of business: $25,000, $40,000, $42,000, $44,000,
$48,000, $50,000. The interest rate is 8% per year.
In cells A1 through A7, enter 100,000, followed by the six annual incomes.
Present value: NPV(8%,A2:A7) A1
The cell A1 value is already a present value. Any year with a zero cash fl ow must have a 0 entered
to ensure a correct result.
PMT (Payments)
Calculates equivalent periodic amounts based on present value and/or future value at a constant
interest rate.
PMT(rate, nper, pv, fv, type)
rate
nper
pv
fv
type
Interest rate per compounding period.
Total number of periods.
Present value.
Future value.
(optional entry) Enter 0 for payments due at the end of the
compounding period and 1 if payment is due at the start of the
compounding period. If omitted, 0 is assumed.
Example Jim plans to take a $15,000 loan to buy a new car. The interest rate is 7% per year.
He wants to pay the loan off in 5 years (60 months). What are his monthly payments?
Monthly payments: PMT(7%12,60,15000)
PPMT (Principal Payment)
Calculates the payment on the principal based on uniform payments at a specified interest
rate.
PPMT(rate, per, nper, pv, fv, type)
rate
per
nper
pv
fv
type
Interest rate per compounding period.
Period for which the payment on the principal is required.
Total number of periods.
Present value.
Future value.
(optional entry) Enter 0 for payments that are due at the end of the
compounding period and 1 if payments are due at the start of the
compounding period. If omitted, 0 is assumed.

556 Appendix A Using Spreadsheets and Microsoft Excel ©
Example Jovita is planning to invest $10,000 in equipment which is expected to last 10 years
with no salvage value. The interest rate is 5%. What is the principal payment at the end of
year 4 and year 8?
At the end of year 4: PPMT(5%,4,10,10000)
At the end of year 8: PPMT(5%,8,10,10000)
PV (Present Value)
Calculates the present value of a future series of equal cash flows and a single lump sum in the
last period at a constant interest rate.
PV(rate, nper, pmt, fv, type)
rate
nper
pmt
fv
type
Interest rate per compounding period.
Total number of periods.
Cash flow at regular intervals. Negative numbers represent payments
(cash outflows), and positive numbers represent income.
Future value or cash balance at the end of the last period.
(optional entry) Enter 0 if payments are due at the end of the
compounding period and 1 if payments are due at the start of each
compounding period. If omitted, 0 is assumed.
There are two primary differences between the PV function and the NPV function: PV allows for
end or beginning of period cash flows, and PV requires that all amounts have the same value,
whereas they may vary for the NPV function.
Example Jose is considering leasing a car for $300 a month for 3 years (36 months). After the
36-month lease, he can purchase the car for $12,000. Using an interest rate of 8% per year, find
the present value of this option.
Present value: PV(8%12,36,300,12000)
Note the minus signs on the pmt and fv amounts.
RAND (Random Number)
Returns an evenly distributed number that is (1) 0 and 1; (2) 0 and 100; or (3) between
two specified numbers.
RAND() for range 0 to 1
RAND()*100 for range 0 to 100
RAND()*(ba)a
a minimum integer to be generated
b maximum integer to be generated
for range a to b
The Excel function RANDBETWEEN(a,b) may also be used to obtain a random number between
two values.
Example Grace needs random numbers between 5 and 10 with 3 digits after the decimal. What
is the Excel function? Here a 5 and b 10.
Random number: RAND()*5 5
Example Randi wants to generate random numbers between the limits of 10 and 25. What
is the Excel function? The minimum and maximum values are a 10 and b 25, so b a
25 (10) 35.
Random number: RAND()*35 10

A.3 Excel Functions Important to Engineering Economy 557
RATE (Interest Rate)
Calculates the interest rate per compounding period for a series of payments or incomes.
RATE(nper, pmt, pv, fv, type, guess)
nper
pmt
pv
fv
type
guess
Total number of periods.
Payment amount made each compounding period.
Present value.
Future value (not including the pmt amount).
(optional entry) Enter 0 for payments due at the end of the
compounding period and 1 if payments are due at the start of each
compounding period. If omitted, 0 is assumed.
(optional entry) To minimize computing time, include a guessed
interest rate. If a value of guess is not specified, the function will
assume a rate of 10%. This function usually converges to a solution if
the rate is between 0% and 100%.
Example Alysha wants to start a savings account at a bank. She will make an initial deposit of
$1000 to open the account and plans to deposit $100 at the beginning of each month. She plans
to do this for the next 3 years (36 months). At the end of 3 years, she wants to have at least $5000.
What is the minimum interest required to achieve this result?
Interest rate: RATE(36,100,1000,5000,1)
SLN (Straight Line Depreciation)
Calculates the straight line depreciation of an asset for a given year.
SLN(cost, salvage, life)
cost
salvage
life
First cost or basis of the asset.
Salvage value.
Depreciation life.
Example Maria purchased a printing machine for $100,000. The machine has an allowed depreciation
life of 8 years and an estimated salvage value of $15,000. What is the depreciation
each year?
Depreciation: SLN(100000,15000,8)
SYD (Sum-of-Years-Digits Depreciation)
Calculates the sum-of-years-digits depreciation of an asset for a given year.
SYD(cost, salvage, life, period)
cost
salvage
life
period
First cost or basis of the asset.
Salvage value.
Depreciation life.
The year for which the depreciation is sought.
Example Jack bought equipment for $100,000 that has a depreciation life of 10 years. The
salvage value is $10,000. What is the depreciation for year 1 and year 9?
Depreciation for year 1: SYD(100000,10000,10,1)
Depreciation for year 9: SYD(100000,10000,10,9)
VDB (Variable Declining Balance)
Calculates the depreciation using the declining balance method with a switch to straight line
depreciation in the year in which straight line has a larger depreciation amount. This function

558 Appendix A Using Spreadsheets and Microsoft Excel ©
automatically implements the switch from DB to SL depreciation, unless specifically instructed
to not switch.
VDB (cost, salvage, life, start_period, end_period, factor, no_switch)
cost
salvage
life
start_period
end_period
factor
no_switch
First cost of the asset.
Salvage value.
Depreciation life.
First period for depreciation to be calculated.
Last period for depreciation to be calculated.
(optional entry) If omitted, the function will use the double
declining rate of 2 n , or twice the straight line rate. Other entries
define the declining balance method, for example, 1.5 for 150%
declining balance.
(optional entry) If omitted or entered as FALSE, the function will
switch from declining balance to straight line depreciation when
the latter is greater than DB depreciation. If entered as TRUE, the
function will not switch to SL depreciation at any time during the
depreciation life.
Example Newly purchased equipment with a first cost of $300,000 has a depreciable life of
10 years with no salvage value. Calculate the 175% declining balance depreciation for the first year
and the ninth year if switching to SL depreciation is acceptable and if switching is not permitted.
Depreciation for first year, with switching: VDB(300000,0,10,0,1,1.75)
Depreciation for ninth year, with switching: VDB(300000,0,10,8,9,1.75)
Depreciation for first year, no switching: VDB(300000,0,10,0,1,1.75,TRUE)
Depreciation for ninth year, no switching: VDB(300000,0,10,8,9,1.75,TRUE)
VDB (for MACRS Depreciation)
The VDB function can be adapted to generate the MACRS annual depreciation amount, when the
start_period and end_period are replaced with the MAX and MIN functions, respectively. As
above, the factor option should be entered if other than DDB rates start the MACRS depreciation.
The VDB format is
VDB(cost,0,life,MAX(0,t1.5),MIN(life,t0.5),factor)
Example Determine the MACRS depreciation for year 4 for a $350,000 asset that has a 20%
salvage value and a MACRS recovery period of 3 years. D 4 $25,926 is the display.
Depreciation for year 4: VDB(350000,0,3,MAX(0,41.5),MIN(3,40.5),2)
Example Find the MACRS depreciation in year 16 for a $350,000-asset with a recovery
period of n 15 years. The optional factor 1.5 is required here, since MACRS starts with
150% DB for n 15-year and 20-year recovery periods. D 16 $10,334.
Depreciation for year 16: VDB(350000,0,15,MAX(0,161.5),MIN(15,160.5),1.5)
Other Functions
There are numerous additional financial functions available on Excel, as well as engineering,
mathematics, trigonometry, statistics, data and time, logical, and information functions. These
can be viewed by clicking the Formulas tab on the Excel toolbar.
A.4 Goal Seek—A Tool for Breakeven and
Sensitivity Analysis
Goal Seek is found on the Excel toolbar labeled Data, followed by What-if Analysis. This tool
changes the value in a specific cell based on a numerical value in another (changing) cell as input
by the user. It is a good tool for sensitivity analysis, breakeven analysis, and “what if?” questions

A.5 Solver—An Optimizing Tool for Capital Budgeting, Breakeven, and Sensitivity Analysis 559
Figure A–3
Goal Seek template used
to specify a cell, a value,
and the changing cell.
B$5
B$5500
Figure A–4
Use of Goal Seek to determine an annual cash flow to increase the rate of return.
when no constraint relations or inequalities are needed. The initial Goal Seek template is pictured
in Figure A–3 . One of the cells (set or changing cell) must contain an equation or spreadsheet
function that uses the other cell to determine a numeric value. Only a single cell can be identified
as the changing cell; however, this limitation can be avoided by using equations rather than specific
numerical inputs in any additional cells also to be changed. This is demonstrated below.
Example A new asset will cost $25,000, generate an annual cash flow of $6000 over its 5-year
life, and have an estimated $500 salvage value. The rate of return using the IRR function is
6.94%. Determine the annual cash flow necessary to raise the return to 10% per year.
Figure A–4 (top left) shows the cash flows and return displayed using the function
IRR(B4:B9) prior to the use of Goal Seek. Note that the initial $6000 is input in cell B5, but
other years’ cash flows are input as equations that refer to B5. The $500 salvage is added for the
last year. This format allows Goal Seek to change only cell B5 while making the other cash flows
have the same value. The tool finds the required cash flow of $6506 to approximate the 10% per
year return. The Goal Seek Status inset indicates that a solution is found. Clicking OK saves all
changed cells; clicking Cancel returns to the original values.
A.5 Solver—An Optimizing Tool for Capital Budgeting,
Breakeven, and Sensitivity Analysis
Solver is a powerful spreadsheet tool to change the value in multiple (one or more) cells based on
the value in a specific (target) cell. It is excellent when solving a capital budgeting problem to
select from independent projects where budget constraints are present. (Section 12.4 details this
application.) The initial Solver template is shown in Figure A–5 .

560 Appendix A Using Spreadsheets and Microsoft Excel ©
Figure A–5
Solver template used to
specify optimization in a
target cell, multiple
changing cells, and
constraint relations.
Set Target Cell box. Enter a cell reference or name. The target cell itself must contain a formula
or function. The value in the cell can be maximized (Max), minimized (Min), or restricted
to a specified value (Value of).
By Changing Cells box. Enter the cell reference for each cell to be adjusted, using commas
between nonadjacent cells. Each cell must be directly or indirectly related to the target cell.
Solver proposes a value for the changing cell based on input provided about the target cell. The
Guess button will list all possible changing cells related to the target cell.
Subject to the Constraints box. Enter any constraints that may apply, for example,
$C$1 $50,000. Integer and binary variables are determined in this box.
Options box. Choices here allow the user to specify various parameters of the solution: maximum
time and number of iterations allowed, the precision and tolerance of the values determined,
and the convergence requirements as the final solution is determined. Also, linear and
nonlinear model assumptions can be set here. If integer or binary variables are involved, the
tolerance option must be set to a small number, say, 0.0001. This is especially important for the
binary variables when selecting from independent projects (Chapter 12). If tolerance remains at
the default value of 5%, a project may be incorrectly included in the solution set at a very low
level.
Solver Results box. This appears after Solve is clicked and a solution appears. It is possible,
of course, that no solution can be found for the scenario described. It is possible to update the
spreadsheet by clicking Keep Solver Solution, or return to the original entries using Restore
Original Values.
A.6 Error Messages
If Excel is unable to complete a formula or function computation, an error message is displayed.
Some of the common messages are:
#DIV0! Requires division by zero.
#NA Refers to a value that is not available.
#NAME? Uses a name that Excel doesn’t recognize.
#NULL! Specifies an invalid intersection of two areas.
#NUM! Uses a number incorrectly.
#REF! Refers to a cell that is not valid.
#VALUE! Uses an invalid argument or operand.
##### Produces a result, or includes a constant numeric value,
that is too long to fit in the cell. (Widen the column.)

APPENDIX B
BASICS OF ACCOUNTING
REPORTS AND BUSINESS RATIOS
This appendix provides a fundamental description of financial statements. The documents discussed
here will assist in reviewing or understanding basic financial statements and in gathering
information useful in an engineering economy study.
B.1 The Balance Sheet
The fiscal year and the tax year are defined identically for a corporation or an individual—
12 months in length. The fiscal year (FY) is commonly not the calendar year (CY) for a corporation.
The U.S. government uses October through September as its FY. For example, October
2011 through September 2012 is FY2012. The fiscal or tax year is always the calendar year for
an individual citizen.
At the end of each fiscal year, a company publishes a balance sheet. A sample balance sheet
for JAGBA Corporation is presented in Table B–1 . This is a yearly presentation of the state of the
firm at a particular time, for example, May 31, 2012; however, a balance sheet is also usually
prepared quarterly and monthly. Three main categories are used.
Assets. This section is a summary of all resources owned by or owed to the company. There
are two main classes of assets. Current assets represent shorter-lived working capital (cash,
accounts receivable, etc.), which is more easily converted to cash, usually within 1 year. Longerlived
assets are referred to as fi xed assets (land, equipment, etc.). Conversion of these holdings
to cash in a short time would require a major corporate reorientation.
Liabilities. This section is a summary of all fi nancial obligations (debts, mortgages, loans,
etc.) of a corporation. Bond indebtedness is included here.
Net worth. Also called owner’s equity, this section provides a summary of the financial
value of ownership, including stocks issued and earnings retained by the corporation.
TABLE B–1
Sample Balance Sheet
Assets
JAGBA CORPORATION
Balance Sheet
May 31, 2012
Liabilities
Current
Cash $10,500 Accounts payable $19,700
Accounts receivable 18,700 Dividends payable 7,000
Interest accrued receivable 500 Long-term notes payable 16,000
Inventories 52,000 Bonds payable 20,000
Total current assets $81,700 Total liabilities $62,700
Fixed
Net Worth
Land $25,000 Common stock $275,000
Building and equipment 438,000 Preferred stock 100,000
Less: Depreciation Retained earnings 25,000
allowance $82,000 356,000
Total fixed assets 381,000 Total net worth 400,000
Total assets $462,700 Total liabilities and net worth $462,700

562 Appendix B Basics of Accounting Reports and Business Ratios
The balance sheet is constructed using the relation
Assets liabilities net worth
In Table B–1 each major category is further divided into standard subcategories. For example,
current assets is comprised of cash, accounts receivable, etc. Each subdivision has a specific interpretation,
such as accounts receivable, which represents all money owed to the company by its
customers.
B.2 Income Statement and Cost of Goods
Sold Statement
A second important financial statement is the income statement ( Table B–2 ). The income statement
summarizes the profits or losses of the corporation for a stated period of time. Income statements
always accompany balance sheets. The major categories of an income statement are
Revenues . This includes all sales and interest revenue that the company has received in the
past accounting period.
Expenses. This is a summary of all expenses (operating and others, including taxes) for
the period. Some expense amounts are itemized in other statements, for example, cost of
goods sold.
The final result of an income statement is the net profit after taxes (NPAT), or NOPAT (O for
operating), the amount used in Chapter 17, Sections 17.1 and 17.7. The income statement, published
at the same time as the balance sheet, uses the basic equation
Revenues expenses profit (or loss)
The cost of goods sold is an important accounting term. It represents the net cost of producing
the product marketed by the firm. Cost of goods sold may also be called factory cost. A statement
of the cost of goods sold, such as that shown in Table B–3 , is useful in determining exactly how
much it costs to make a particular product over a stated time period, usually a year. The total of
the cost of goods sold statement is entered as an expense item on the income statement. This total
is determined using the relations
Cost of goods sold prime cost indirect cost
Prime cost direct materials direct labor
[B.1]
TABLE B–2
Sample Income Statement
JAGBA CORPORATION
Income Statement
Year Ended May 31, 2012
Revenues
Sales $505,000
Interest revenue 3,500
Total revenues $508,500
Expenses
Cost of goods sold (from Table B–3) $290,000
Selling 28,000
Administrative 35,000
Other 12,000
Total expenses 365,000
Income before taxes 143,500
Taxes for year 64,575
Net profit after taxes (NPAT) $ 78,925

B.3 Business Ratios 563
TABLE B–3
Sample Cost of Goods Sold Statement
JAGBA CORPORATION
Statement of Cost of Goods Sold
Year Ended May 31, 2012
Materials
Inventory June 1, 2011 $ 54,000
Purchases during year 174,500
Total $228,500
Less: Inventory May 31, 2012 50,000
Cost of materials $178,500
Direct labor 110,000
Prime cost 288,500
Indirect costs 7,000
Factory cost 295,500
Less: Increase in finished goods inventory during year 5,500
Cost of goods sold (into Table B–2) $290,000
Indirect costs include all indirect and overhead charges made to a product, process, or cost center.
Indirect cost allocation methods are discussed in Chapter 15.
B.3 Business Ratios
Accountants, financial analysts, and engineering economists frequently utilize business ratio
analysis to evaluate the financial health (status) of a company over time and in relation to industry
norms. Because the engineering economist must continually communicate with others, she or
he should have a basic understanding of several ratios. For comparison purposes, it is necessary
to compute the ratios for several companies in the same industry. Industrywide median ratio values
are published annually by firms such as Dun and Bradstreet in Industry Norms and Key Business
Ratios. The ratios are classified according to their role in measuring the corporation.
Solvency ratios. Assess ability to meet short-term and long-term financial obligations.
Efficiency ratios. Measure management’s ability to use and control assets.
Profitability ratios. Evaluate the ability to earn a return for the owners of the corporation.
Numerical data for several important ratios are discussed here and are extracted from the JAGBA
balance sheet and income statement, Tables B–1 and B–2.
Current Ratio This ratio is utilized to analyze the company’s working capital condition. It is
defined as
Current ratio ———————
current assets
current liabilities
Current liabilities include all short-term debts, such as accounts and dividends payable. Note that
only balance sheet data are utilized in the current ratio; that is, no association with revenues or
expenses is made. For the balance sheet of Table B–1 , current liabilities amount to $19,700
$7000 $26,700 and
Current ratio ———
81,700
26,700 3.06
Since current liabilities are those debts payable in the next year, the current ratio value of 3.06
means that the current assets would cover short-term debts approximately 3 times. Current ratio
values of 2 to 3 are common.
The current ratio assumes that the working capital invested in inventory can be converted to
cash quite rapidly. Often, however, a better idea of a company’s immediate financial position can
be obtained by using the acid test ratio.

564 Appendix B Basics of Accounting Reports and Business Ratios
Acid Test Ratio (Quick Ratio) This ratio is
quick assets
Acid-test ratio ———————
current liabilities
————————————
current assets inventories
current liabilities
It is meaningful for the emergency situation when the firm must cover short-term debts using its
readily convertible assets. For JAGBA Corporation,
81,700 52,000
Acid test ratio ——————— 1.11
26,700
Comparison of this and the current ratio shows that approximately 2 times the current debt of the
company is invested in inventories. However, an acid test ratio of approximately 1.0 is generally
regarded as a strong current position, regardless of the amount of assets in inventories.
Debt Ratio This ratio is a measure of financial strength since it is defined as
Debt ratio ——————
total liabilities
total assets
For JAGBA Corporation,
Debt ratio ————
62,700
462,700 0.136
JAGBA is 13.6% creditor-owned and 86.4% stockholder-owned. A debt ratio in the range of 20%
or less usually indicates a sound financial condition, with little fear of forced reorganization because
of unpaid liabilities. However, a company with virtually no debts, that is, one with a very low debt
ratio, may not have a promising future, because of its inexperience in dealing with short-term and
long-term debt financing. The debt-equity (D-E) mix is another measure of financial strength.
Return on Sales Ratio This often quoted ratio indicates the profit margin for the company. It
is defined as
net profit
Return on sales ————
net sales (100%)
Net profit is the after-tax value from the income statement. This ratio measures profit earned per
sales dollar and indicates how well the corporation can sustain adverse conditions over time, such
as falling prices, rising costs, and declining sales. For JAGBA Corporation,
Return on sales ————
78,925 (100%) 15.6%
505,000
Corporations may point to small return on sales ratios, say, 2.5% to 4.0%, as indications of sagging
economic conditions. In truth, for a relatively large-volume, high-turnover business, an income
ratio of 3% is quite healthy. Of course, a steadily decreasing ratio indicates rising company
expenses, which absorb net profit after taxes.
Return on Assets Ratio This is the key indicator of profitability since it evaluates the ability
of the corporation to transfer assets into operating profit. The definition and value for
JAGBA are
net profit
Return on assets —————
total assets (100%)
————
78,925 (100%) 17.1%
462,700
Efficient use of assets indicates that the company should earn a high return, while low returns
usually accompany lower values of this ratio compared to the industry group ratios.
Inventory Turnover Ratio Two different ratios are used here. They both indicate the number
of times the average inventory value passes through the operations of the company. If turnover of
inventory to net sales is desired, the formula is
Net sales to inventory ————————
net sales
average inventory

B.3 Business Ratios 565
where average inventory is the figure recorded in the balance sheet. For JAGBA Corporation this
ratio is
Net sales to inventory ————
505,000
52,000 9.71
This means that the average value of the inventory has been sold 9.71 times during the year.
Values of this ratio vary greatly from one industry to another.
If inventory turnover is related to cost of goods sold, the ratio to use is
cost of goods sold
Cost of goods sold to inventory ————————
average inventory
Now, average inventory is computed as the average of the beginning and ending inventory values
in the statement of cost of goods sold. This ratio is commonly used as a measure of the inventory
turnover rate in manufacturing companies. It varies with industries, but management likes to see
it remain relatively constant as business increases. For JAGBA, using the values in Table B–3 ,
290,000
Cost of goods sold to inventory ————————— 5.58
1_ (54,000 50,000)
2
There are, of course, many other ratios to use in various circumstances; however, the ones
presented here are commonly used by both accountants and economic analysts.
EXAMPLE B.1
Sample values for financial ratios or percentages of four industry sectors are presented below.
Compare the corresponding JAGBA Corporation values with these norms, and comment on
differences and similarities.
Ratio or
Percentage
Motor Vehicles
and Parts
Manufacturing
336105*
Air
Transportation
(Medium-Sized)
481000*
Industrial
Machinery
Manufacturing
333200*
Home
Furnishings
442000*
Current ratio 2.4 0.4 2.2 2.6
Quick ratio 1.6 0.3 1.5 1.2
Debt ratio 59.3% 96.8% 49.1% 52.4%
Return on
assets
40.9% 8.1% 8.0% 5.1%
*North American Industry Classification System (NAICS) code for this industry sector.
SOURCE: L. Troy, Almanac of Business and Industrial Financial Ratios, CCH, Wolters Kluwer, USA.
Solution
It is not correct to compare ratios for one company with indexes in different industries, that is,
with indexes for different NAICS codes. So the comparison below is for illustration purposes
only. The corresponding values for JAGBA are
Current ratio 3.06
Quick ratio 1.11
Debt ratio 13.5%
Return on assets 17.1%
JAGBA has a current ratio larger than all four of these industries, since 3.06 indicates it can
cover current liabilities 3 times compared with 2.6 and much less in the case of the “average”
air transportation corporation. JAGBA has a significantly lower debt ratio than that of any of
the sample industries, so it is likely more financially sound. Return on assets, which is a measure
of ability to turn assets into profitability, is not as high at JAGBA as motor vehicles, but
JAGBA competes well with the other industry sectors.
To make a fair comparison of JAGBA ratios with other values, it is necessary to have norm
values for its industry type as well as ratio values for other corporations in the same NAICS
category and about the same size in total assets. Corporate assets are classified in categories by
$100,000 units, such as 100 to 250, 1001 to 5000, over 250,000, etc.

APPENDIX C
CODE OF ETHICS FOR ENGINEERS
Source: National Society of Professional Engineers (www.nspe.org).

Appendix C Code of Ethics for Engineers 567
Code of Ethics for Engineers
Preamble
Engineering is an important and learned profession. As members of this
profession, engineers are expected to exhibit the highest standards of honesty
and integrity. Engineering has a direct and vital impact on the quality of life for
all people. Accordingly, the services provided by engineers require honesty,
impartiality, fairness, and equity, and must be dedicated to the protection of the
public health, safety, and welfare. Engineers must perform under a standard of
professional behavior that requires adherence to the highest principles of ethical
conduct.
I. Fundamental Canons
Engineers, in the fulfillment of their professional duties, shall:
1. Hold paramount the safety, health, and welfare of the public.
2. Perform services only in areas of their competence.
3. Issue public statements only in an objective and truthful manner.
4. Act for each employer or client as faithful agents or trustees.
5. Avoid deceptive acts.
6. Conduct themselves honorably, responsibly, ethically, and
lawfully so as to enhance the honor, reputation, and usefulness
of the profession.
II. Rules of Practice
1. Engineers shall hold paramount the safety, health, and welfare
of the public.
a. If engineers’ judgment is overruled under circumstances that
endanger life or property, they shall notify their employer or client
and such other authority as may be appropriate.
b. Engineers shall approve only those engineering documents that are
in conformity with applicable standards.
c. Engineers shall not reveal facts, data, or information without the
prior consent of the client or employer except as authorized or
required by law or this Code.
d. Engineers shall not permit the use of their name or associate in
business ventures with any person or firm that they believe is
engaged in fraudulent or dishonest enterprise.
e. Engineers shall not aid or abet the unlawful practice of engineering
by a person or firm.
f. Engineers having knowledge of any alleged violation of this Code
shall report thereon to appropriate professional bodies and, when
relevant, also to public authorities, and cooperate with the proper
authorities in furnishing such information or assistance as may be
required.
2. Engineers shall perform services only in the areas of their
competence.
a. Engineers shall undertake assignments only when qualified by
education or experience in the specific technical fields involved.
b. Engineers shall not affix their signatures to any plans or documents
dealing with subject matter in which they lack competence, nor to
any plan or document not prepared under their direction and
control.
c. Engineers may accept assignments and assume responsibility for
coordination of an entire project and sign and seal the engineering
documents for the entire project, provided that each technical
segment is signed and sealed only by the qualified engineers who
prepared the segment.
3. Engineers shall issue public statements only in an objective and
truthful manner.
a. Engineers shall be objective and truthful in professional reports,
statements, or testimony. They shall include all relevant and
pertinent information in such reports, statements, or testimony,
which should bear the date indicating when it was current.
b. Engineers may express publicly technical opinions that are founded
upon knowledge of the facts and competence in the subject matter.
c. Engineers shall issue no statements, criticisms, or arguments on
technical matters that are inspired or paid for by interested parties,
unless they have prefaced their comments by explicitly identifying
the interested parties on whose behalf they are speaking, and by
revealing the existence of any interest the engineers may have in the
matters.
4. Engineers shall act for each employer or client as faithful agents or
trustees.
a. Engineers shall disclose all known or potential conflicts of interest
that could influence or appear to influence their judgment or the
quality of their services.
b. Engineers shall not accept compensation, financial or otherwise,
from more than one party for services on the same project, or for
services pertaining to the same project, unless the circumstances are
fully disclosed and agreed to by all interested parties.
c. Engineers shall not solicit or accept financial or other valuable
consideration, directly or indirectly, from outside agents in
connection with the work for which they are responsible.
d. Engineers in public service as members, advisors, or employees
of a governmental or quasi-governmental body or department shall
not participate in decisions with respect to services solicited or
provided by them or their organizations in private or public
engineering practice.
e. Engineers shall not solicit or accept a contract from a governmental
body on which a principal or officer of their organization serves as
a member.
5. Engineers shall avoid deceptive acts.
a. Engineers shall not falsify their qualifications or permit
misrepresentation of their or their associates’ qualifications. They
shall not misrepresent or exaggerate their responsibility in or for the
subject matter of prior assignments. Brochures or other
presentations incident to the solicitation of employment shall not
misrepresent pertinent facts concerning employers, employees,
associates, joint venturers, or past accomplishments.
b. Engineers shall not offer, give, solicit, or receive, either directly or
indirectly, any contribution to influence the award of a contract by
public authority, or which may be reasonably construed by the
public as having the effect or intent of influencing the awarding of a
contract. They shall not offer any gift or other valuable
consideration in order to secure work. They shall not pay a
commission, percentage, or brokerage fee in order to secure work,
except to a bona fide employee or bona fide established commercial
or marketing agencies retained by them.
III. Professional Obligations
1. Engineers shall be guided in all their relations by the highest standards
of honesty and integrity.
a. Engineers shall acknowledge their errors and shall not distort or
alter the facts.
b. Engineers shall advise their clients or employers when they believe
a project will not be successful.
c. Engineers shall not accept outside employment to the detriment of
their regular work or interest. Before accepting any outside
engineering employment, they will notify their employers.
d. Engineers shall not attempt to attract an engineer from another
employer by false or misleading pretenses.
e. Engineers shall not promote their own interest at the expense of the
dignity and integrity of the profession.
2. Engineers shall at all times strive to serve the public interest.
a. Engineers are encouraged to participate in civic affairs; career
guidance for youths; and work for the advancement of the safety,
health, and well-being of their community.
b. Engineers shall not complete, sign, or seal plans and/or
specifications that are not in conformity with applicable engineering
standards. If the client or employer insists on such unprofessional
conduct, they shall notify the proper authorities and withdraw from
further service on the project.
c. Engineers are encouraged to extend public knowledge and
appreciation of engineering and its achievements.
d. Engineers are encouraged to adhere to the principles of sustainable
development 1 in order to protect the environment for future
generations.

568 Appendix C Code of Ethics for Engineers
3. Engineers shall avoid all conduct or practice that deceives the public.
a. Engineers shall avoid the use of statements containing a material
misrepresentation of fact or omitting a material fact.
b. Consistent with the foregoing, engineers may advertise for
recruitment of personnel.
c. Consistent with the foregoing, engineers may prepare articles for
the lay or technical press, but such articles shall not imply credit to
the author for work performed by others.
4. Engineers shall not disclose, without consent, confidential information
concerning the business affairs or technical processes of any present or
former client or employer, or public body on which they serve.
a. Engineers shall not, without the consent of all interested parties,
promote or arrange for new employment or practice in connection
with a specific project for which the engineer has gained particular
and specialized knowledge.
b. Engineers shall not, without the consent of all interested parties,
participate in or represent an adversary interest in connection with a
specific project or proceeding in which the engineer has gained
particular specialized knowledge on behalf of a former client or
employer.
5. Engineers shall not be influenced in their professional duties by
conflicting interests.
a. Engineers shall not accept financial or other considerations,
including free engineering designs, from material or equipment
suppliers for specifying their product.
b. Engineers shall not accept commissions or allowances, directly or
indirectly, from contractors or other parties dealing with clients or
employers of the engineer in connection with work for which the
engineer is responsible.
6. Engineers shall not attempt to obtain employment or advancement or
professional engagements by untruthfully criticizing other engineers,
or by other improper or questionable methods.
a. Engineers shall not request, propose, or accept a commission on a
contingent basis under circumstances in which their judgment may
be compromised.
b. Engineers in salaried positions shall accept part-time engineering
work only to the extent consistent with policies of the employer and
in accordance with ethical considerations.
c. Engineers shall not, without consent, use equipment, supplies,
laboratory, or office facilities of an employer to carry on outside
private practice.
7. Engineers shall not attempt to injure, maliciously or falsely, directly
or indirectly, the professional reputation, prospects, practice, or
employment of other engineers. Engineers who believe others are
guilty of unethical or illegal practice shall present such information
to the proper authority for action.
a. Engineers in private practice shall not review the work of another
engineer for the same client, except with the knowledge of such
engineer, or unless the connection of such engineer with the work
has been terminated.
b. Engineers in governmental, industrial, or educational employ are
entitled to review and evaluate the work of other engineers when so
required by their employment duties.
c. Engineers in sales or industrial employ are entitled to make
engineering comparisons of represented products with products of
other suppliers.
8. Engineers shall accept personal responsibility for their professional
activities, provided, however, that engineers may seek indemnification
for services arising out of their practice for other than gross
negligence, where the engineer’s interests cannot otherwise be
protected.
a. Engineers shall conform with state registration laws in the practice
of engineering.
b. Engineers shall not use association with a nonengineer, a
corporation, or partnership as a “cloak” for unethical acts.
9. Engineers shall give credit for engineering work to those to whom
credit is due, and will recognize the proprietary interests of others.
a. Engineers shall, whenever possible, name the person or persons
who may be individually responsible for designs, inventions,
writings, or other accomplishments.
b. Engineers using designs supplied by a client recognize that the
designs remain the property of the client and may not be duplicated
by the engineer for others without express permission.
c. Engineers, before undertaking work for others in connection with
which the engineer may make improvements, plans, designs,
inventions, or other records that may justify copyrights or patents,
should enter into a positive agreement regarding ownership.
d. Engineers’ designs, data, records, and notes referring exclusively to
an employer’s work are the employer’s property. The employer
should indemnify the engineer for use of the information for any
purpose other than the original purpose.
e. Engineers shall continue their professional development throughout
their careers and should keep current in their specialty fields by
engaging in professional practice, participating in continuing
education courses, reading in the technical literature, and attending
professional meetings and seminars.
Footnote 1 “Sustainable development” is the challenge of meeting human
needs for natural resources, industrial products, energy, food,
transportation, shelter, and effective waste management while
conserving and protecting environmental quality and the natural
resource base essential for future development.
As Revised July 2007
“By order of the United States District Court for the District of Columbia,
former Section 11(c) of the NSPE Code of Ethics prohibiting competitive
bidding, and all policy statements, opinions, rulings or other guidelines
interpreting its scope, have been rescinded as unlawfully interfering with the
legal right of engineers, protected under the antitrust laws, to provide price
information to prospective clients; accordingly, nothing contained in the NSPE
Code of Ethics, policy statements, opinions, rulings or other guidelines prohibits
the submission of price quotations or competitive bids for engineering services
at any time or in any amount.”
Statement by NSPE Executive Committee
In order to correct misunderstandings which have been indicated in some
instances since the issuance of the Supreme Court decision and the entry of the
Final Judgment, it is noted that in its decision of April 25, 1978, the Supreme
Court of the United States declared: “The Sherman Act does not require
competitive bidding.”
It is further noted that as made clear in the Supreme Court decision:
1. Engineers and firms may individually refuse to bid for engineering services.
2. Clients are not required to seek bids for engineering services.
3. Federal, state, and local laws governing procedures to procure engineering
services are not affected, and remain in full force and effect.
4. State societies and local chapters are free to actively and aggressively seek
legislation for professional selection and negotiation procedures by public
agencies.
5. State registration board rules of professional conduct, including rules
prohibiting competitive bidding for engineering services, are not affected and
remain in full force and effect. State registration boards with authority to
adopt rules of professional conduct may adopt rules governing procedures to
obtain engineering services.
6. As noted by the Supreme Court, “nothing in the judgment prevents NSPE and
its members from attempting to influence governmental action . . .”
Note: In regard to the question of application of the Code to corporations vis-a-vis real persons, business form or type should not negate nor
influence conformance of individuals to the Code. The Code deals with professional services, which services must be performed by real
persons. Real persons in turn establish and implement policies within business structures. The Code is clearly written to apply to the Engineer,
and it is incumbent on members of NSPE to endeavor to live up to its provisions. This applies to all pertinent sections of the Code.
1420 King Street
Alexandria, Virginia 22314-2794
703/684-2800 • Fax:703/836-4875
www.nspe.org
Publication date as revised: July 2007 • Publication #1102

APPENDIX D
ALTERNATE METHODS FOR
EQUIVALENCE CALCULATIONS
Throughout the text, engineering economy factor formulas, tabulated factor values, or built-in
spreadsheet functions have been used to obtain a value of P, F, A, i , or n . Because of advances in
programmable and scientific calculators, many of the equivalence computations can be performed
without the use of tables or spreadsheets, but rather with a handheld calculator. An overview
of the possibilities is presented here.
Alternatively, the recognition that all equivalence calculations involve geometric series can
likewise remove the need for tabulated values or spreadsheet functions. From the summation of
the series, it is possible to perform calculator-based computations to obtain P, F , or A values. A
brief introduction to this technique is presented in Section D.2.
D.1 Using Programmable Calculators
A basic way to calculate one parameter, given the other four, is to use a calculator that allows a
present worth relation to be encoded. The software can then solve for any one of the parameters,
when the remaining four are entered. For example, consider a PW relation in which all five parameters
are included.
A ( P/A,i,n ) F ( P/F,i,n ) P 0
The A, P, and F values can be positive (cash inflow) or negative (cash outflow) or zero, as long
as there is at least one value with each sign. The interest rate i can be coded for entry as a percent
or decimal. When the unknown variable is identified, the calculator’s software can solve the
equation for zero, thus providing the answer.
This is the approach taken by relatively simple scientific calculators, such as the Hewlett-
Packard (HP) scientific series, for example, HP 33s. By substituting the formulas for the factors,
the actual relation entered into the calculator is
n
A [
1 (1 i/100)
————————
i/100 ] F [1 (i/100)] n P 0
The HP calculator uses a slightly different symbol set than we have used thus far. The initial investment
is called B rather than P , and the equal uniform amount is termed P rather than A . Once
entered, the relation can be solved for any one variable, given values for the other four.
Another example that offers freedom from the spreadsheet and tables is an engineering calculator
that has the same functions built in as those on a spreadsheet to determine P, F, A, i , or n . An
example of this higher level is Texas Instrument’s TI-Nspire series. The functions are basically
the same as those on a spreadsheet. They are clustered under the heading of tvm (time value of
money) functions. For example, the tvmPV function format is
where
tvmPV(n,i,Pmt,FV,PpY,CpY,PmtAt)
n number of periods
i annual interest rate as a percent
Pmt equal uniform periodic amount A
FV future amount F
PpY payments per year (optional; default is 1)
CpY compounding periods per year (optional; default is 1)
PmtAt beginning- or end-of-period payments (optional; default is 0 end)

570 Appendix D Alternate Methods for Equivalence Calculations
It is easy to understand why it is possible to do a lot on a calculator with relatively wellbehaved
cash flow series. As the series become more complex, it is necessary to move to a spreadsheet
for speed and versatility. However, the use of tables or factor formulas is not necessary.
D.2 Using the Summation of a Geometric Series
A geometric progression is a series of n terms with a common ratio or base r . If c is a constant for
each term, the series is written in the form
cr a cr a 1 ⋅ ⋅ ⋅ cr n c
r j
ja
The sum S of a geometric series adds the terms using the closed-end form
jn
S r n1 r a
————
r 1
[D.1]
Ristroph
1
and others have explained how the recognition that equivalence computations are
simple applications of geometric series can be used to determine F, P , and A using Equation
[D.1] and a simple handheld calculator with exponentiation capability.
Before explaining how to apply this approach, we define the base r as follows when a future
worth F or present worth P is sought.
To find F :
To find P :
r 1 i
r (1 i) 1
A very familiar application of geometric series is the determination of the equivalent future worth
F in year n for a single present worth amount P in year 0. This is the same as using a geometric
series of only one term. As shown in Figure D–1 , if P $100, n 10 years, and i 10% per
year, when r 1 i ,
F P (1 i )
n
P ( r )
n
100(1.1) 10
100(2.5937)
$259.37
This is identical to using the tabulated value (or formula) for the F/P factor.
F P ( F/P,i,n ) 100( F/P ,10%,10) 100(2.5937)
$259.37
Figure D–2 shows a uniform annual series A $100 for 10 years. To calculate F , we can move
each A value forward to year 10. Place a subscript j on each A value to indicate the year of occurrence,
and determine F for each A j value.
F A 1 (1 i ) 9 A 2 (1 i ) 8 ⋅ ⋅ ⋅ A 9 (1 i )
1 A 10 (1 i ) 0
Figure D–1
Future worth of a single
amount in year 0.
0
F = ?
i = 10%
1 2 3 4 5 6 7 8 9 10 Year
1 2 3 4 5 6 7 8 9 10 Times
compounded
P = $100
1
J. H. Ristroph, “Engineering Economics: Time for New Directions?” Proceedings , ASEE
Annual Conference, Austin, TX, June 2009.

D.2 Using the Summation of a Geometric Series 571
F = ?
A $100
i = 10%
0 1 2 3 4 5 6 7 8 9 10
Year
Times
9 8 7 6 5 4 3 2 1
compounded
0
Figure D–2
Future worth of an A series from year 1 through year 10.
P = ?
i = 10%
0 1 2 3 4 5 6 7 8 9 10
Year
A $100
5 6 7 8 9 10
Figure D–3
Present worth of a shifted uniform series.
Times
discounted
Note that the first value A 1 is compounded 9 times, not 10. With r 1 i , the geometric series
and its sum S (from Equation [D.1]) can be developed. Removing the subscript on A ,
j9
F = A [ r 9 r 8 ⋅ ⋅ ⋅ r 1 r 0 j
] A r
j0
S ————
r10 r 0
r 1 (1.1)10 1
————— 15.9374
0.1
The F value is the same whether using the geometric series sum or the tabulated F/A factor.
F 100( S ) 100(15.9374) $1593.74
F 100( F/A ,10%,10) 100(15.9374) $1593.74
As a final demonstration of the geometric series approach to equivalence computations, consider
the shifted cash flow series in Figure D–3 . The A series is present from years 5 through 10,
and the P value is sought. If the tables are used, the solution is
P A ( P/A ,10%,6)( P/F ,10%,4) = 100(4.3553)(0.6830)
$297.47

572 Appendix D Alternate Methods for Equivalence Calculations
As shown in the figure, this is a geometric series with the first A value discounted 5 years, therefore,
a 5. The last A term is discounted 10 years, making n 10. Since P is sought, the geometric
series base is r (1 i )
1
. Again using Equation [D.1] for the summation, we have
j10
P A
[ r
] j j5
100 [ r 11 r 5
————
r 1 ] 100 [ (1/1.1) 11 (1/1.1) 5
————————
100 [ (0.9091) 11 (0.9091) 5
——————————
0.9091 1 ] 100(2.9747)
$297.47
(1/1.1) 1 ]
It is possible to develop similar relations to handle arithmetic and geometric series, conversions
from P to A , and vice versa. Proponents of this approach point out the use of standard
mathematical notation; the removal of a need to derive any factors; no need to remember the
placement of P, F, and A values based on factor formula development; and the easy use of a calculator
to determine the equivalence relations. As with the use of programmable calculators, the
technique is excellent for well-behaved and reasonably complex series. When the series become
quite involved, or when sensitivity analysis is required to reach an economic decision, it may
be beneficial to use a spreadsheet. This often helps in performing sidebar and ancillary calculations
that assist in the understanding of the problem, not just the math computations necessary to
obtain an answer. However, once again, the use of tables and factors is not necessary.

APPENDIX E
GLOSSARY OF CONCEPTS
AND TERMS
E.1 Important Concepts and Guidelines
The following elements of engineering economy are identified throughout the text in the margin
by this checkmark and a title below it. The numbers in parentheses indicate chapters where the
concept or guideline is introduced or essential to obtaining a correct solution.
Title
Time Value of Money It is a fact that money makes money. This concept explains the change
in the amount of money over time for both owned and borrowed funds. (1)
Economic Equivalence A combination of time value of money and interest rate that makes
different sums of money at different times have equal economic value . (1)
Cash Flow The flow of money into and out of a company, project, or activity. Revenues are
cash infl ows and carry a positive (+) sign; expenses are outfl ows and carry a negative (−) sign.
If only costs are involved, the − sign may be omitted, e.g., benefit/cost (B/C) analysis. (1, 9)
End-of-Period Convention To simplify calculations, cash flows (revenues and costs) are
assumed to occur at the end of a time period . An interest period or fiscal period is commonly
1 year . A half-year convention is often used in depreciation calculations. (1)
Cost of Capital The interest rate incurred to obtain capital investment funds. COC is usually
a weighted average that involves the cost of debt capital (loans, bonds, and mortgages) and equity
capital (stocks and retained earnings). (1, 10)
Minimum Attractive Rate of Return (MARR) A reasonable rate of return established for
the evaluation of an economic alternative. Also called the hurdle rate, MARR is based on cost of
capital, market trend, risk, etc. The inequality ROR ≥ MARR > COC is correct for an economically
viable project. (1, 10)
Opportunity Cost A forgone opportunity caused by the inability to pursue a project. Numerically,
it is the largest rate of return of all the projects not funded due to the lack of capital
funds. Stated differently, it is the ROR of the first project rejected because of unavailability of
funds. (1, 10)
Nominal or Effective Interest Rate ( r or i ) A nominal interest rate does not include any
compounding ; for example, 1% per month is the same as nominal 12% per year. Effective interest
rate is the actual rate over a period of time because compounding is imputed ; for example, 1% per
month, compounded monthly, is an effective 12.683% per year. Inflation or deflation is not considered.
(4)
Placement of Present Worth ( P ; PW) In applying the ( P / A , i %, n ) factor, P or PW is always
located one interest period (year) prior to the fi rst A amount . The A or AW is a series of equal,
end-of-period cash flows for n consecutive periods, expressed as money per time (say,
$/year; /year). (2, 3)
Placement of Future Worth ( F ; FW) In applying the ( F / A , i %, n ) factor, F or FW is always
located at the end of the last interest period (year) of the A series . (2, 3)

574 Appendix E Glossary of Concepts and Terms
Placement of Gradient Present Worth ( P G ; P g ) The ( P / G , i %, n ) factor for an arithmetic
gradient finds the P G of only the gradient series 2 years prior to the first appearance of the constant
gradient G. The base amount A is treated separately from the gradient series.
The ( P /A,g,i,n) factor for a geometric gradient determines P g for the gradient and initial amount
A 1 two years prior to the appearance of the first gradient amount. The initial amount A 1 is included
in the value of P g . (2, 3)
Equal-Service Requirement Identical capacity of all alternatives operating over the
same amount of time is mandated by the equal-service requirement. Estimated costs and revenues
for equal service must be evaluated. PW analysis requires evaluation over the same
number of years (periods) using the LCM (least common multiple) of lives; AW analysis is
performed over one life cycle. Further, equal service assumes that all costs and revenues rise
and fall in accordance with the overall rate of inflation or deflation over the total time period
of the evaluation. (5, 6, 8)
LCM or Study Period To select from mutually exclusive alternatives under the equal-service
requirement for PW computations, use the LCM of lives with repurchase(s) as necessary. For a
stated study period (planning horizon), evaluate cash flows only over this period , neglecting any
beyond this time; estimated market values at termination of the study period are the salvage
v alues. (5, 6, 11)
Salvage/Market Value Expected trade-in, market, or scrap value at the end of the estimated
life or the study period . In a replacement study, the defender’s estimated market value at the end
of a year is considered its “first cost” at the beginning of the next year. MACRS depreciation always
reduces the book value to a salvage of zero. (6, 11)
Do Nothing The DN alternative is always an option, unless one of the defined alternatives
must be selected. DN is status quo; it generates no new costs, revenues, or savings . (5)
Revenue or Cost Alternative Revenue alternatives have costs and revenues estimated; savings
are considered negative costs and carry a + sign. Incremental evaluation requires comparison
with DN for revenue alternatives. Cost (or service) alternatives have only costs estimated;
revenues and savings are assumed equal between alternatives. (5)
Rate of Return An interest rate that equates a PW or AW relation to zero . Also defined as the
rate on the unpaid balance of borrowed money, or rate earned on the unrecovered balance of an
investment such that the last cash fl ow brings the balance exactly to zero. (7, 8)
Project Evaluation For a specifi ed MARR, determine a measure of worth for net cash flow
series over the life or study period. Guidelines for a single project to be economically justified at
the MARR (or discount rate) follow. (5, 6, 7, 9, 17)
Present worth: If PW ≥ 0 Annual worth: If AW ≥ 0
Future worth: If FW ≥ 0 Rate of return: If i * ≥ MARR
Benefit/cost: If B/C ≥ 1.0 Profitability index: If PI ≥ 1.0
ME Alternative Selection For mutually exclusive (select only one) alternatives, compare
two alternatives at a time by determining a measure of worth for the incremental (∆) cash flow
series over the life or study period, adhering to the equal-service requirement. (5, 6, 8, 9, 10, 17)
Present worth or annual worth: Find PW or AW values at MARR; select numerically l argest
(least negative or most positive).
Rate of return: Order by initial cost , perform pairwise ∆ i * comparison; if ∆ i * ≥ MARR,
select larger cost alternative; continue until one remains.
Benefit/cost: Order by total equivalent cost , perform pairwise ∆B/C comparison; if
∆B/C ≥ 1.0, select larger cost alternative; continue until one remains.

E.1 Important Concepts and Guidelines 575
Cost-effectiveness ratio: For service sector alternatives; order by effectiveness measure ;
perform pairwise ∆C/E comparison using dominance ; select from nondominated alternatives
without exceeding budget.
Independent Project Selection No comparison between projects; only against DN. Calculate
a measure of worth and select using the guidelines below. (5, 6, 8, 9, 12)
Present worth or annual worth: Find PW or AW at MARR; select all projects with PW or
AW ≥ 0.
Rate of return: No incremental comparison; select all projects with overall i * ≥ MARR.
Benefit/cost: No incremental comparison; select all projects with overall B/C ≥ 1.0.
Cost-effectiveness ratio: For service sector projects; no incremental comparison; order by
CER and select projects to not exceed budget.
When a capital budget limit is defined, independent projects are selected using the capital budgeting
process based on PW values . The Solver spreadsheet tool is useful here.
Capital Recovery CR is the equivalent annual amount an asset or system must earn to recover
the initial investment plus a stated rate of return. Numerically, it is the AW value of the initial
investment at a stated rate of return. The salvage value is considered in CR calculations. (6)
Economic Service Life The ESL is the number of years n at which the total AW of costs ,
including salvage and AOC, is at its minimum, considering all the years the asset may provide
service. (11)
Sunk Cost Capital (money) that is lost and cannot be recovered. Sunk costs are not included
when making decisions about the future. They should be handled using tax laws and write-off
allowances, not the economic study. (11)
Inflation Expressed as a percentage per time (% per year), it is an increase in the amount
of money required to purchase the same amount of goods or services over time . Inflation
occurs when the value of a currency decreases. Economic evaluations are performed using
either a market (inflation-adjusted) interest rate or an inflation-free rate (constant-value
terms). (1, 14)
Breakeven For a single project, the value of a parameter that makes two elements equal,
e.g., sales necessary to equate revenues and costs. For two alternatives, breakeven is the value
of a common variable at which the two are equally acceptable. Breakeven analysis is fundamental
to make-buy decisions, replacement studies, payback analysis, sensitivity analysis,
breakeven ROR analysis, and many others. The Goal Seek spreadsheet tool is useful in breakeven
analysis. (8, 13)
Payback Period Amount of time n before recovery of the initial capital investment is expected.
Payback with i > 0 or simple payback at i = 0 is useful for preliminary or screening analysis to
determine if a full PW, AW, or ROR analysis is needed. (13)
Direct / Indirect Costs Direct costs are primarily human labor, machines, and materials associated
with a product, process, system, or service. Indirect costs, which include support functions,
utilities, management, legal, taxes, and the like, are more difficult to associate with a
specific product or process. (15)
Value Added Activities have added worth to a product or service from the perspective of a
consumer, owner, or investor who is willing to pay more for an enhanced value. (17)
Sensitivity Analysis Determination of how a measure of worth is affected by changes in
estimated values of a parameter over a stated range. Parameters may be any cost factor, revenue,
life, salvage value, inflation rate, etc. (18)

576 Appendix E Glossary of Concepts and Terms
Risk Variation from an expected, desirable, or predicted value that may be detrimental to the
product, process, or system. Risk represents an absence of or deviation from certainty . Probability
estimates of variation (values) help evaluate risk and uncertainty using statistics and simulation.
(10, 18, 19, 20)
E.2 Symbols and Terms
This section identifies and defines the common terms and their symbols used throughout the text.
The numbers in parentheses indicate sections where the term is introduced and used in various
applications.
Term Symbol Description
Annual amount or
worth
A or AW Equivalent uniform annual worth of all cash inflows and
outflows over estimated life (1.5, 6.l).
Annual operating cost AOC Estimated annual costs to maintain and support an
alternative (1.3).
Benefit/cost ratio B/C Ratio of a project’s benefits to costs expressed in PW, AW,
or FW terms (9.2).
Book value BV Remaining capital investment in an asset after depreciation
is accounted for (16.1).
Breakeven point Q BE Quantity at which revenues and costs are equal, or two
alternatives are equivalent (13.1).
Capital budget b Amount of money available for capital investment projects
(12.1).
Capital recovery CR or A Equivalent annual cost of owning an asset plus the required
return on the initial investment (6.2).
Capitalized cost CC or P Present worth of an alternative that will last forever (or a
long time) (5.5).
Cash flow CF Actual cash amounts that are receipts (inflow) and disbursements
(outflow) (1.6).
Cash flow before or
after taxes
Compounding
frequency
CFBT or
CFAT
m
Cash flow amount before relevant taxes or after taxes are
applied (17.2).
Number of times interest is compounded per period (year)
(4.1).
Cost-effectiveness ratio CER Ratio of equivalent cost to effectiveness measure to evaluate
service sector projects (9.5).
Cost estimating relationships
C 2 or C T
Relations that use design variables and changing costs over
time to estimate current and future costs (15.3–4).
Cost of capital WACC Interest rate paid for the use of capital funds; includes both
debt and equity funds. For debt and equity considered, it is
weighted average cost of capital (1.9, 10.2).
Debt-equity mix D-E Percentages of debt and equity investment capital used by a
corporation (10.2).
Depreciation D Reduction in the value of assets using specific models and
rules; there are book and tax depreciation methods (16.1).
Depreciation rate d t Annual rate for reducing the value of assets using different
depreciation methods (16.1).
Economic service life ESL or n Number of years at which the AW of costs is a minimum
(11.2).
Effectiveness measure E A nonmonetary measure used in the cost-effectiveness ratio
for service sector projects (9.5).

E.2 Symbols and Terms 577
Term Symbol Description
Expected value
( average)
X – , , or E(X) Long-run expected average if a random variable is sampled
many times (18.3, 19.4).
Expenses, operating OE All corporate costs incurred in transacting business (17.1).
First cost P Total initial cost—purchase, construction, setup, etc.
(1.3, 16.1).
Future amount or worth F or FW Amount at some future date considering time value of
money (1.5, 5.4).
Gradient, arithmetic G Uniform change (+ or –) in cash flow each time period
(2.5).
Gradient, geometric g Constant rate of change (+ or –) each time period (2.6).
Gross income GI Income from all sources for corporations or individuals
(17.1).
Inflation rate f Rate that reflects changes in the value of a currency over
time (14.1).
Interest I Amount earned or paid over time based on an initial
amount and interest rate (1.4).
Interest rate i or r Interest expressed as a percentage of the original amount per
time period; nominal (r) and effective (i) rates (1.4, 4.1).
Interest rate, inflationadjusted
i f Interest rate adjusted to take inflation into account (14.1).
Life (estimated) n Number of years or periods over which an alternative or
asset will be used; the evaluation time (1.5).
Life-cycle cost LCC Evaluation of costs for a system over all stages: feasibility
to design to phaseout (6.5).
Measure of worth Varies Value, such as PW, AW, i*, used to judge economic
viability (1.1).
Minimum attractive
rate of return
MARR
Minimum value of the rate of return for an alternative to be
financially viable (1.9, 10.1).
Modified ROR i' or MIRR Unique ROR when a reinvestment rate i i and external borrowing
rate i b are applied to multiple-rate cash flows (7.5).
Net cash flow NCF Resulting, actual amount of cash that flows in or out during
a time period (1.6).
Net operating income NOI Difference between gross income and operating expenses
(17.1).
Net operating profit
after taxes
NOPAT or
NPAT
Amount remaining after taxes are removed from taxable
income (17.1).
Net present value NPV Another name for the present worth, PW.
Payback period n p Number of years to recover the initial investment and a
stated rate of return (13.3).
Present amount
or worth
P or PW Amount of money at the current time or a time denoted as
present (1.5, 5.2).
Probability distribution P(X) Distribution of probability over different values of a
v ariable (19.2).
Profitability index PI Ratio of PW of net cash flows to initial investment used for
revenue projects; rewritten modified B/C ratio (9.2, 12.5).
Random variable X Parameter or characteristic that can take on any one of
several values; discrete and continuous (19.2).
Rate of return i* or ROR Compound interest rate on unpaid or unrecovered balances
such that the final amount results in a zero balance (7.1).

578 Appendix E Glossary of Concepts and Terms
Term Symbol Description
Recovery period n Number of years to completely depreciate an asset (16.1).
Return on invested
capital
i'' or ROIC Unique ROR when a reinvestment rate i i is applied to
multiple-rate cash flows (7.5).
Salvage/market value S or MV Expected trade-in or market value when an asset is traded
or disposed of (6.2, 11.1, 16.1).
Standard deviation s or σ Measure of dispersion or spread about the expected value
or average (19.4).
Study period n Specified number of years over which an evaluation takes
place (5.3, 11.5).
Taxable income TI Amount upon which income taxes are based (17.1).
Tax rate T Decimal rate, usually graduated, used to calculate corporate
or individual taxes (17.1).
Tax rate, effective T e Single-figure tax rate incorporating several rates and bases
(17.1).
Time t Indicator for a time period (1.7).
Unadjusted basis B Depreciable amount of first cost, delivery, and installation
costs of an asset (18.1).
Value added EVA Economic value added reflects net profit after taxes (NPAT)
after removing cost of invested capital during the year
(17.7).
Value-added tax VAT An indirect consumption tax collected at each stage of
production/distribution process; different from a sales tax
paid by end user at purchase time (17.9).

REFERENCE MATERIALS
Textbooks on Related Topics
Blank, L. T., and A. Tarquin: Basics of Engineering Economy, McGraw-Hill, New York, 2008.
Bowman, M. S.: Applied Economic Analysis for Technologists, Engineers, and Managers, 2d ed., Pearson
Prentice Hall, Upper Saddle River, NJ, 2003.
Bussey, L. E., and T. G. Eschenbach: The Economic Analysis of Industrial Projects , 2d ed., Pearson
Prentice Hall, Upper Saddle River, NJ, 1992.
Canada, J. R., W. G. Sullivan, D. J. Kulonda, and J. A. White: Capital Investment Analysis for Engineering
and Management , 3d ed., Pearson Prentice Hall, Upper Saddle River, NJ, 2005.
Collier, C. A., and C. R. Glagola: Engineering Economic and Cost Analysis , 3d ed., Pearson Prentice Hall,
Upper Saddle River, NJ, 1999.
Cushman, R. F., and M. Loulakis: Design-Build Contracting Handbook, 2d ed., Aspen Publishers, New
York, 2001.
Eschenbach, T. G.: Engineering Economy: Applying Theory to Practice , 3d ed., Oxford University Press,
New York, 2010.
Fabrycky, W. J., G. J. Thuesen, and D. Verma: Economic Decision Analysis , 3d ed., Pearson Prentice Hall,
Upper Saddle River, NJ, 1998.
Fraser, N. M., E. M. Jewkes, I. Bernhardt, and M. Tajima: Engineering Economics in Canada, Pearson
Prentice Hall, Upper Saddle River, NJ, 2006.
Hartman, J. C.: Engineering Economy and the Decision Making Process , Pearson Prentice Hall, Upper
Saddle River, NJ, 2007.
Levy, S. M.: Build, Operate, Transfer: Paving the Way for Tomorrow’s Infrastructure , John Wiley & Sons,
New York, 1996.
Newnan, D. G., J. P. Lavelle, and T. G. Eschenbach: Engineering Economic Analysis , 10th ed., Oxford
University Press, New York, 2009.
Ostwald, P. F.: Construction Cost Analysis and Estimating , Pearson Prentice Hall, Upper Saddle River,
NJ, 2001.
Ostwald, P. F., and T. S. McLaren: Cost Analysis and Estimating for Engineering and Management ,
Pearson Prentice Hall, Upper Saddle River, NJ, 2004.
Park, C. S.: Contemporary Engineering Economics , 5th ed., Pearson Prentice Hall, Upper Saddle River,
NJ, 2011.
Park, C. S.: Fundamentals of Engineering Economics , 2d ed., Pearson Prentice Hall, Upper Saddle River,
NJ, 2008.
Peurifoy, R. L., and G. D. Oberlender: Estimating Construction Costs, 5th ed., McGraw-Hill, New York,
2002.
Riggs, J. L., D. D. Bedworth, and S. U. Randhawa: Engineering Economics , 4th ed., McGraw-Hill,
New York, 1996.
Stewart, R. D., R. M. Wyskida, and J. D. Johannes: Cost Estimator’s Reference Manual , 2d ed., John
Wiley & Sons, New York, 1995.
Sullivan, W. G., E. M. Wicks, and C. P. Koelling: Engineering Economy , 15th ed., Pearson Prentice Hall,
Upper Saddle River, NJ, 2011.
Thuesen, G. J., and W. J. Fabrycky: Engineering Economy , 9th ed., Pearson Prentice Hall, Upper Saddle
River, NJ, 2001.
White, J. A., K. E. Case, D. B. Pratt, and M. H. Agee: Principles of Engineering Economic Analysis ,
5th ed., John Wiley & Sons, New York, 2010.
Using Excel 2007
Gottfried, B. S.: Spreadsheet Tools for Engineers Using Excel ® 2007 , McGraw-Hill, New York, 2010.
Materials on Engineering Ethics
Harris, C. E., M. S. Pritchard, and M. J. Rabins: Engineering Ethics: Concepts and Cases , 4th ed.,
Wadsworth Cengage Learning, Belmont, CA, 2009.
Martin, M. W., and R. Schinzinger: Introduction to Engineering Ethics , 2d ed., McGraw-Hill,
New York, 2010.

580 Reference Materials
Websites
Construction cost estimation index: www.construction.com
The Economist : www.economist.com
For this textbook: www.mhhe.com/blank
Plant cost estimation index: www.che.com/pci
Revenue Canada: www.cra.gc.ca
U.S. Internal Revenue Service: www.irs.gov
Wall Street Journal : www.online.wsj.com
U.S. Government Publications (available at
www.irs.gov )
Corporations , Publication 544, Internal Revenue Service, GPO, Washington, DC, annually.
Sales and Other Dispositions of Assets , Publication 542, Internal Revenue Service, GPO, Washington,
DC, annually.
Your Federal Income Tax , Publication 17, Internal Revenue Service, GPO, Washington, DC, annually.
Selected Journals and Other Publications
The Engineering Economist, joint publication of ASEE and IIE, published by Taylor and Francis,
Philadelphia, quarterly.
Harvard Business Review , Harvard University Press, Boston, bimonthly.
Journal of Finance , American Finance Association, published by John Wiley & Sons, New York,
bimonthly.

Compound Interest Factor Tables 581
0.25% TABLE 1 Discrete Cash Flow: Compound Interest Factors 0.25%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.0025 0.9975 1.00000 1.0000 1.00250 0.9975
2 1.0050 0.9950 0.49938 2.0025 0.50188 1.9925 0.9950 0.4994
3 1.0075 0.9925 0.33250 3.0075 0.33500 2.9851 2.9801 0.9983
4 1.0100 0.9901 0.24906 4.0150 0.25156 3.9751 5.9503 1.4969
5 1.0126 0.9876 0.19900 5.0251 0.20150 4.9627 9.9007 1.9950
6 1.0151 0.9851 0.16563 6.0376 0.16813 5.9478 14.8263 2.4927
7 1.0176 0.9827 0.14179 7.0527 0.14429 6.9305 20.7223 2.9900
8 1.0202 0.9802 0.12391 8.0704 0.12641 7.9107 27.5839 3.4869
9 1.0227 0.9778 0.11000 9.0905 0.11250 8.8885 35.4061 3.9834
10 1.0253 0.9753 0.09888 10.1133 0.10138 9.8639 44.1842 4.4794
11 1.0278 0.9729 0.08978 11.1385 0.09228 10.8368 53.9133 4.9750
12 1.0304 0.9705 0.08219 12.1664 0.08469 11.8073 64.5886 5.4702
13 1.0330 0.9681 0.07578 13.1968 0.07828 12.7753 76.2053 5.9650
14 1.0356 0.9656 0.07028 14.2298 0.07278 13.7410 88.7587 6.4594
15 1.0382 0.9632 0.06551 15.2654 0.06801 14.7042 102.2441 6.9534
16 1.0408 0.9608 0.06134 16.3035 0.06384 15.6650 116.6567 7.4469
17 1.0434 0.9584 0.05766 17.3443 0.06016 16.6235 131.9917 7.9401
18 1.0460 0.9561 0.05438 18.3876 0.05688 17.5795 148.2446 8.4328
19 1.0486 0.9537 0.05146 19.4336 0.05396 18.5332 165.4106 8.9251
20 1.0512 0.9513 0.04882 20.4822 0.05132 19.4845 183.4851 9.4170
21 1.0538 0.9489 0.04644 21.5334 0.04894 20.4334 202.4634 9.9085
22 1.0565 0.9466 0.04427 22.5872 0.04677 21.3800 222.3410 10.3995
23 1.0591 0.9442 0.04229 23.6437 0.04479 22.3241 243.1131 10.8901
24 1.0618 0.9418 0.04048 24.7028 0.04298 23.2660 264.7753 11.3804
25 1.0644 0.9395 0.03881 25.7646 0.04131 24.2055 287.3230 11.8702
26 1.0671 0.9371 0.03727 26.8290 0.03977 25.1426 310.7516 12.3596
27 1.0697 0.9348 0.03585 27.8961 0.03835 26.0774 335.0566 12.8485
28 1.0724 0.9325 0.03452 28.9658 0.03702 27.0099 360.2334 13.3371
29 1.0751 0.9301 0.03329 30.0382 0.03579 27.9400 386.2776 13.8252
30 1.0778 0.9278 0.03214 31.1133 0.03464 28.8679 413.1847 14.3130
36 1.0941 0.9140 0.02658 37.6206 0.02908 34.3865 592.4988 17.2306
40 1.1050 0.9050 0.02380 42.0132 0.02630 38.0199 728.7399 19.1673
48 1.1273 0.8871 0.01963 50.9312 0.02213 45.1787 1040.06 23.0209
50 1.1330 0.8826 0.01880 53.1887 0.02130 46.9462 1125.78 23.9802
52 1.1386 0.8782 0.01803 55.4575 0.02053 48.7048 1214.59 24.9377
55 1.1472 0.8717 0.01698 58.8819 0.01948 51.3264 1353.53 26.3710
60 1.1616 0.8609 0.01547 64.6467 0.01797 55.6524 1600.08 28.7514
72 1.1969 0.8355 0.01269 78.7794 0.01519 65.8169 2265.56 34.4221
75 1.2059 0.8292 0.01214 82.3792 0.01464 68.3108 2447.61 35.8305
84 1.2334 0.8108 0.01071 93.3419 0.01321 75.6813 3029.76 40.0331
90 1.2520 0.7987 0.00992 100.7885 0.01242 80.5038 3446.87 42.8162
96 1.2709 0.7869 0.00923 108.3474 0.01173 85.2546 3886.28 45.5844
100 1.2836 0.7790 0.00881 113.4500 0.01131 88.3825 4191.24 47.4216
108 1.3095 0.7636 0.00808 123.8093 0.01058 94.5453 4829.01 51.0762
120 1.3494 0.7411 0.00716 139.7414 0.00966 103.5618 5852.11 56.5084
132 1.3904 0.7192 0.00640 156.1582 0.00890 112.3121 6950.01 61.8813
144 1.4327 0.6980 0.00578 173.0743 0.00828 120.8041 8117.41 67.1949
240 1.8208 0.5492 0.00305 328.3020 0.00555 180.3109 19399 107.5863
360 2.4568 0.4070 0.00172 582.7369 0.00422 237.1894 36264 152.8902
480 3.3151 0.3016 0.00108 926.0595 0.00358 279.3418 53821 192.6699

582 Compound Interest Factor Tables
0.5% TABLE 2 Discrete Cash Flow: Compound Interest Factors 0.5%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.0050 0.9950 1.00000 1.0000 1.00500 0.9950
2 1.0100 0.9901 0.49875 2.0050 0.50375 1.9851 0.9901 0.4988
3 1.0151 0.9851 0.33167 3.0150 0.33667 2.9702 2.9604 0.9967
4 1.0202 0.9802 0.24813 4.0301 0.25313 3.9505 5.9011 1.4938
5 1.0253 0.9754 0.19801 5.0503 0.20301 4.9259 9.8026 1.9900
6 1.0304 0.9705 0.16460 6.0755 0.16960 5.8964 14.6552 2.4855
7 1.0355 0.9657 0.14073 7.1059 0.14573 6.8621 20.4493 2.9801
8 1.0407 0.9609 0.12283 8.1414 0.12783 7.8230 27.1755 3.4738
9 1.0459 0.9561 0.10891 9.1821 0.11391 8.7791 34.8244 3.9668
10 1.0511 0.9513 0.09777 10.2280 0.10277 9.7304 43.3865 4.4589
11 1.0564 0.9466 0.08866 11.2792 0.09366 10.6770 52.8526 4.9501
12 1.0617 0.9419 0.08107 12.3356 0.08607 11.6189 63.2136 5.4406
13 1.0670 0.9372 0.07464 13.3972 0.07964 12.5562 74.4602 5.9302
14 1.0723 0.9326 0.06914 14.4642 0.07414 13.4887 86.5835 6.4190
15 1.0777 0.9279 0.06436 15.5365 0.06936 14.4166 99.5743 6.9069
16 1.0831 0.9233 0.06019 16.6142 0.06519 15.3399 113.4238 7.3940
17 1.0885 0.9187 0.05651 17.6973 0.06151 16.2586 128.1231 7.8803
18 1.0939 0.9141 0.05323 18.7858 0.05823 17.1728 143.6634 8.3658
19 1.0994 0.9096 0.05030 19.8797 0.05530 18.0824 160.0360 8.8504
20 1.1049 0.9051 0.04767 20.9791 0.05267 18.9874 177.2322 9.3342
21 1.1104 0.9006 0.04528 22.0840 0.05028 19.8880 195.2434 9.8172
22 1.1160 0.8961 0.04311 23.1944 0.04811 20.7841 214.0611 10.2993
23 1.1216 0.8916 0.04113 24.3104 0.04613 21.6757 233.6768 10.7806
24 1.1272 0.8872 0.03932 25.4320 0.04432 22.5629 254.0820 11.2611
25 1.1328 0.8828 0.03765 26.5591 0.04265 23.4456 275.2686 11.7407
26 1.1385 0.8784 0.03611 27.6919 0.04111 24.3240 297.2281 12.2195
27 1.1442 0.8740 0.03469 28.8304 0.03969 25.1980 319.9523 12.6975
28 1.1499 0.8697 0.03336 29.9745 0.03836 26.0677 343.4332 13.1747
29 1.1556 0.8653 0.03213 31.1244 0.03713 26.9330 367.6625 13.6510
30 1.1614 0.8610 0.03098 32.2800 0.03598 27.7941 392.6324 14.1265
36 1.1967 0.8356 0.02542 39.3361 0.03042 32.8710 557.5598 16.9621
40 1.2208 0.8191 0.02265 44.1588 0.02765 36.1722 681.3347 18.8359
48 1.2705 0.7871 0.01849 54.0978 0.02349 42.5803 959.9188 22.5437
50 1.2832 0.7793 0.01765 56.6452 0.02265 44.1428 1035.70 23.4624
52 1.2961 0.7716 0.01689 59.2180 0.02189 45.6897 1113.82 24.3778
55 1.3156 0.7601 0.01584 63.1258 0.02084 47.9814 1235.27 25.7447
60 1.3489 0.7414 0.01433 69.7700 0.01933 51.7256 1448.65 28.0064
72 1.4320 0.6983 0.01157 86.4089 0.01657 60.3395 2012.35 33.3504
75 1.4536 0.6879 0.01102 90.7265 0.01602 62.4136 2163.75 34.6679
84 1.5204 0.6577 0.00961 104.0739 0.01461 68.4530 2640.66 38.5763
90 1.5666 0.6383 0.00883 113.3109 0.01383 72.3313 2976.08 41.1451
96 1.6141 0.6195 0.00814 122.8285 0.01314 76.0952 3324.18 43.6845
100 1.6467 0.6073 0.00773 129.3337 0.01273 78.5426 3562.79 45.3613
108 1.7137 0.5835 0.00701 142.7399 0.01201 83.2934 4054.37 48.6758
120 1.8194 0.5496 0.00610 163.8793 0.01110 90.0735 4823.51 53.5508
132 1.9316 0.5177 0.00537 186.3226 0.01037 96.4596 5624.59 58.3103
144 2.0508 0.4876 0.00476 210.1502 0.00976 102.4747 6451.31 62.9551
240 3.3102 0.3021 0.00216 462.0409 0.00716 139.5808 13416 96.1131
360 6.0226 0.1660 0.00100 1004.52 0.00600 166.7916 21403 128.3236
480 10.9575 0.0913 0.00050 1991.49 0.00550 181.7476 27588 151.7949

Compound Interest Factor Tables 583
0.75% TABLE 3 Discrete Cash Flow: Compound Interest Factors 0.75%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.0075 0.9926 1.00000 1.0000 1.00750 0.9926
2 1.0151 0.9852 0.49813 2.0075 0.50563 1.9777 0.9852 0.4981
3 1.0227 0.9778 0.33085 3.0226 0.33835 2.9556 2.9408 0.9950
4 1.0303 0.9706 0.24721 4.0452 0.25471 3.9261 5.8525 1.4907
5 1.0381 0.9633 0.19702 5.0756 0.20452 4.8894 9.7058 1.9851
6 1.0459 0.9562 0.16357 6.1136 0.17107 5.8456 14.4866 2.4782
7 1.0537 0.9490 0.13967 7.1595 0.14717 6.7946 20.1808 2.9701
8 1.0616 0.9420 0.12176 8.2132 0.12926 7.7366 26.7747 3.4608
9 1.0696 0.9350 0.10782 9.2748 0.11532 8.6716 34.2544 3.9502
10 1.0776 0.9280 0.09667 10.3443 0.10417 9.5996 42.6064 4.4384
11 1.0857 0.9211 0.08755 11.4219 0.09505 10.5207 51.8174 4.9253
12 1.0938 0.9142 0.07995 12.5076 0.08745 11.4349 61.8740 5.4110
13 1.1020 0.9074 0.07352 13.6014 0.08102 12.3423 72.7632 5.8954
14 1.1103 0.9007 0.06801 14.7034 0.07551 13.2430 84.4720 6.3786
15 1.1186 0.8940 0.06324 15.8137 0.07074 14.1370 96.9876 6.8606
16 1.1270 0.8873 0.05906 16.9323 0.06656 15.0243 110.2973 7.3413
17 1.1354 0.8807 0.05537 18.0593 0.06287 15.9050 124.3887 7.8207
18 1.1440 0.8742 0.05210 19.1947 0.05960 16.7792 139.2494 8.2989
19 1.1525 0.8676 0.04917 20.3387 0.05667 17.6468 154.8671 8.7759
20 1.1612 0.8612 0.04653 21.4912 0.05403 18.5080 171.2297 9.2516
21 1.1699 0.8548 0.04415 22.6524 0.05165 19.3628 188.3253 9.7261
22 1.1787 0.8484 0.04198 23.8223 0.04948 20.2112 206.1420 10.1994
23 1.1875 0.8421 0.04000 25.0010 0.04750 21.0533 224.6682 10.6714
24 1.1964 0.8358 0.03818 26.1885 0.04568 21.8891 243.8923 11.1422
25 1.2054 0.8296 0.03652 27.3849 0.04402 22.7188 263.8029 11.6117
26 1.2144 0.8234 0.03498 28.5903 0.04248 23.5422 284.3888 12.0800
27 1.2235 0.8173 0.03355 29.8047 0.04105 24.3595 305.6387 12.5470
28 1.2327 0.8112 0.03223 31.0282 0.03973 25.1707 327.5416 13.0128
29 1.2420 0.8052 0.03100 32.2609 0.03850 25.9759 350.0867 13.4774
30 1.2513 0.7992 0.02985 33.5029 0.03735 26.7751 373.2631 13.9407
36 1.3086 0.7641 0.02430 41.1527 0.03180 31.4468 524.9924 16.6946
40 1.3483 0.7416 0.02153 46.4465 0.02903 34.4469 637.4693 18.5058
48 1.4314 0.6986 0.01739 57.5207 0.02489 40.1848 886.8404 22.0691
50 1.4530 0.6883 0.01656 60.3943 0.02406 41.5664 953.8486 22.9476
52 1.4748 0.6780 0.01580 63.3111 0.02330 42.9276 1022.59 23.8211
55 1.5083 0.6630 0.01476 67.7688 0.02226 44.9316 1128.79 25.1223
60 1.5657 0.6387 0.01326 75.4241 0.02076 48.1734 1313.52 27.2665
72 1.7126 0.5839 0.01053 95.0070 0.01803 55.4768 1791.25 32.2882
75 1.7514 0.5710 0.00998 100.1833 0.01748 57.2027 1917.22 33.5163
84 1.8732 0.5338 0.00859 116.4269 0.01609 62.1540 2308.13 37.1357
90 1.9591 0.5104 0.00782 127.8790 0.01532 65.2746 2578.00 39.4946
96 2.0489 0.4881 0.00715 139.8562 0.01465 68.2584 2853.94 41.8107
100 2.1111 0.4737 0.00675 148.1445 0.01425 70.1746 3040.75 43.3311
108 2.2411 0.4462 0.00604 165.4832 0.01354 73.8394 3419.90 46.3154
120 2.4514 0.4079 0.00517 193.5143 0.01267 78.9417 3998.56 50.6521
132 2.6813 0.3730 0.00446 224.1748 0.01196 83.6064 4583.57 54.8232
144 2.9328 0.3410 0.00388 257.7116 0.01138 87.8711 5169.58 58.8314
240 6.0092 0.1664 0.00150 667.8869 0.00900 111.1450 9494.12 85.4210
360 14.7306 0.0679 0.00055 1830.74 0.00805 124.2819 13312 107.1145
480 36.1099 0.0277 0.00021 4681.32 0.00771 129.6409 15513 119.6620

584 Compound Interest Factor Tables
1% TABLE 4 Discrete Cash Flow: Compound Interest Factors 1%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.0100 0.9901 1.00000 1.0000 1.01000 0.9901
2 1.0201 0.9803 0.49751 2.0100 0.50751 1.9704 0.9803 0.4975
3 1.0303 0.9706 0.33002 3.0301 0.34002 2.9410 2.9215 0.9934
4 1.0406 0.9610 0.24628 4.0604 0.25628 3.9020 5.8044 1.4876
5 1.0510 0.9515 0.19604 5.1010 0.20604 4.8534 9.6103 1.9801
6 1.0615 0.9420 0.16255 6.1520 0.17255 5.7955 14.3205 2.4710
7 1.0721 0.9327 0.13863 7.2135 0.14863 6.7282 19.9168 2.9602
8 1.0829 0.9235 0.12069 8.2857 0.13069 7.6517 26.3812 3.4478
9 1.0937 0.9143 0.10674 9.3685 0.11674 8.5660 33.6959 3.9337
10 1.1046 0.9053 0.09558 10.4622 0.10558 9.4713 41.8435 4.4179
11 1.1157 0.8963 0.08645 11.5668 0.09645 10.3676 50.8067 4.9005
12 1.1268 0.8874 0.07885 12.6825 0.08885 11.2551 60.5687 5.3815
13 1.1381 0.8787 0.07241 13.8093 0.08241 12.1337 71.1126 5.8607
14 1.1495 0.8700 0.06690 14.9474 0.07690 13.0037 82.4221 6.3384
15 1.1610 0.8613 0.06212 16.0969 0.07212 13.8651 94.4810 6.8143
16 1.1726 0.8528 0.05794 17.2579 0.06794 14.7179 107.2734 7.2886
17 1.1843 0.8444 0.05426 18.4304 0.06426 15.5623 120.7834 7.7613
18 1.1961 0.8360 0.05098 19.6147 0.06098 16.3983 134.9957 8.2323
19 1.2081 0.8277 0.04805 20.8109 0.05805 17.2260 149.8950 8.7017
20 1.2202 0.8195 0.04542 22.0190 0.05542 18.0456 165.4664 9.1694
21 1.2324 0.8114 0.04303 23.2392 0.05303 18.8570 181.6950 9.6354
22 1.2447 0.8034 0.04086 24.4716 0.05086 19.6604 198.5663 10.0998
23 1.2572 0.7954 0.03889 25.7163 0.04889 20.4558 216.0660 10.5626
24 1.2697 0.7876 0.03707 26.9735 0.04707 21.2434 234.1800 11.0237
25 1.2824 0.7798 0.03541 28.2432 0.04541 22.0232 252.8945 11.4831
26 1.2953 0.7720 0.03387 29.5256 0.04387 22.7952 272.1957 11.9409
27 1.3082 0.7644 0.03245 30.8209 0.04245 23.5596 292.0702 12.3971
28 1.3213 0.7568 0.03112 32.1291 0.04112 24.3164 312.5047 12.8516
29 1.3345 0.7493 0.02990 33.4504 0.03990 25.0658 333.4863 13.3044
30 1.3478 0.7419 0.02875 34.7849 0.03875 25.8077 355.0021 13.7557
36 1.4308 0.6989 0.02321 43.0769 0.03321 30.1075 494.6207 16.4285
40 1.4889 0.6717 0.02046 48.8864 0.03046 32.8347 596.8561 18.1776
48 1.6122 0.6203 0.01633 61.2226 0.02633 37.9740 820.1460 21.5976
50 1.6446 0.6080 0.01551 64.4632 0.02551 39.1961 879.4176 22.4363
52 1.6777 0.5961 0.01476 67.7689 0.02476 40.3942 939.9175 23.2686
55 1.7285 0.5785 0.01373 72.8525 0.02373 42.1472 1032.81 24.5049
60 1.8167 0.5504 0.01224 81.6697 0.02224 44.9550 1192.81 26.5333
72 2.0471 0.4885 0.00955 104.7099 0.01955 51.1504 1597.87 31.2386
75 2.1091 0.4741 0.00902 110.9128 0.01902 52.5871 1702.73 32.3793
84 2.3067 0.4335 0.00765 130.6723 0.01765 56.6485 2023.32 35.7170
90 2.4486 0.4084 0.00690 144.8633 0.01690 59.1609 2240.57 37.8724
96 2.5993 0.3847 0.00625 159.9273 0.01625 61.5277 2459.43 39.9727
100 2.7048 0.3697 0.00587 170.4814 0.01587 63.0289 2605.78 41.3426
108 2.9289 0.3414 0.00518 192.8926 0.01518 65.8578 2898.42 44.0103
120 3.3004 0.3030 0.00435 230.0387 0.01435 69.7005 3334.11 47.8349
132 3.7190 0.2689 0.00368 271.8959 0.01368 73.1108 3761.69 51.4520
144 4.1906 0.2386 0.00313 319.0616 0.01313 76.1372 4177.47 54.8676
240 10.8926 0.0918 0.00101 989.2554 0.01101 90.8194 6878.60 75.7393
360 35.9496 0.0278 0.00029 3494.96 0.01029 97.2183 8720.43 89.6995
480 118.6477 0.0084 0.00008 11765 0.01008 99.1572 9511.16 95.9200

Compound Interest Factor Tables 585
1.25% TABLE 5 Discrete Cash Flow: Compound Interest Factors 1.25%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.0125 0.9877 1.00000 1.0000 1.01250 0.9877
2 1.0252 0.9755 0.49680 2.0125 0.50939 1.9631 0.9755 0.4969
3 1.0380 0.9634 0.32920 3.0377 0.34170 2.9265 2.9023 0.9917
4 1.0509 0.9515 0.24536 4.0756 0.25786 3.8781 5.7569 1.4845
5 1.0641 0.9398 0.19506 5.1266 0.20756 4.8178 9.5160 1.9752
6 1.0774 0.9282 0.16153 6.1907 0.17403 5.7460 14.1569 2.4638
7 1.0909 0.9167 0.13759 7.2680 0.15009 6.6627 19.6571 2.9503
8 1.1045 0.9054 0.11963 8.3589 0.13213 7.5681 25.9949 3.4348
9 1.1183 0.8942 0.10567 9.4634 0.11817 8.4623 33.1487 3.9172
10 1.1323 0.8832 0.09450 10.5817 0.10700 9.3455 41.0973 4.3975
11 1.1464 0.8723 0.08537 11.7139 0.09787 10.2178 49.8201 4.8758
12 1.1608 0.8615 0.07776 12.8604 0.09026 11.0793 59.2967 5.3520
13 1.1753 0.8509 0.07132 14.0211 0.08382 11.9302 69.5072 5.8262
14 1.1900 0.8404 0.06581 15.1964 0.07831 12.7706 80.4320 6.2982
15 1.2048 0.8300 0.06103 16.3863 0.07353 13.6005 92.0519 6.7682
16 1.2199 0.8197 0.05685 17.5912 0.06935 14.4203 104.3481 7.2362
17 1.2351 0.8096 0.05316 18.8111 0.06566 15.2299 117.3021 7.7021
18 1.2506 0.7996 0.04988 20.0462 0.06238 16.0295 130.8958 8.1659
19 1.2662 0.7898 0.04696 21.2968 0.05946 16.8193 145.1115 8.6277
20 1.2820 0.7800 0.04432 22.5630 0.05682 17.5993 159.9316 9.0874
21 1.2981 0.7704 0.04194 23.8450 0.05444 18.3697 175.3392 9.5450
22 1.3143 0.7609 0.03977 25.1431 0.05227 19.1306 191.3174 10.0006
23 1.3307 0.7515 0.03780 26.4574 0.05030 19.8820 207.8499 10.4542
24 1.3474 0.7422 0.03599 27.7881 0.04849 20.6242 224.9204 10.9056
25 1.3642 0.7330 0.03432 29.1354 0.04682 21.3573 242.5132 11.3551
26 1.3812 0.7240 0.03279 30.4996 0.04529 22.0813 260.6128 11.8024
27 1.3985 0.7150 0.03137 31.8809 0.04387 22.7963 279.2040 12.2478
28 1.4160 0.7062 0.03005 33.2794 0.04255 23.5025 298.2719 12.6911
29 1.4337 0.6975 0.02882 34.6954 0.04132 24.2000 317.8019 13.1323
30 1.4516 0.6889 0.02768 36.1291 0.04018 24.8889 337.7797 13.5715
36 1.5639 0.6394 0.02217 45.1155 0.03467 28.8473 466.2830 16.1639
40 1.6436 0.6084 0.01942 51.4896 0.03192 31.3269 559.2320 17.8515
48 1.8154 0.5509 0.01533 65.2284 0.02783 35.9315 759.2296 21.1299
50 1.8610 0.5373 0.01452 68.8818 0.02702 37.0129 811.6738 21.9295
52 1.9078 0.5242 0.01377 72.6271 0.02627 38.0677 864.9409 22.7211
55 1.9803 0.5050 0.01275 78.4225 0.02525 39.6017 946.2277 23.8936
60 2.1072 0.4746 0.01129 88.5745 0.02379 42.0346 1084.84 25.8083
72 2.4459 0.4088 0.00865 115.6736 0.02115 47.2925 1428.46 30.2047
75 2.5388 0.3939 0.00812 123.1035 0.02062 48.4890 1515.79 31.2605
84 2.8391 0.3522 0.00680 147.1290 0.01930 51.8222 1778.84 34.3258
90 3.0588 0.3269 0.00607 164.7050 0.01857 53.8461 1953.83 36.2855
96 3.2955 0.3034 0.00545 183.6411 0.01795 55.7246 2127.52 38.1793
100 3.4634 0.2887 0.00507 197.0723 0.01757 56.9013 2242.24 39.4058
108 3.8253 0.2614 0.00442 226.0226 0.01692 59.0865 2468.26 41.7737
120 4.4402 0.2252 0.00363 275.2171 0.01613 61.9828 2796.57 45.1184
132 5.1540 0.1940 0.00301 332.3198 0.01551 64.4781 3109.35 48.2234
144 5.9825 0.1672 0.00251 398.6021 0.01501 66.6277 3404.61 51.0990
240 19.7155 0.0507 0.00067 1497.24 0.01317 75.9423 5101.53 67.1764
360 87.5410 0.0114 0.00014 6923.28 0.01264 79.0861 5997.90 75.8401
480 388.7007 0.0026 0.00003 31016 0.01253 79.7942 6284.74 78.7619

586 Compound Interest Factor Tables
1.5% TABLE 6 Discrete Cash Flow: Compound Interest Factors 1.5%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.0150 0.9852 1.00000 1.0000 1.01500 0.9852
2 1.0302 0.9707 0.49628 2.0150 0.51128 1.9559 0.9707 0.4963
3 1.0457 0.9563 0.32838 3.0452 0.34338 2.9122 2.8833 0.9901
4 1.0614 0.9422 0.24444 4.0909 0.25944 3.8544 5.7098 1.4814
5 1.0773 0.9283 0.19409 5.1523 0.20909 4.7826 9.4229 1.9702
6 1.0934 0.9145 0.16053 6.2296 0.17553 5.6972 13.9956 2.4566
7 1.1098 0.9010 0.13656 7.3230 0.15156 6.5982 19.4018 2.9405
8 1.1265 0.8877 0.11858 8.4328 0.13358 7.4859 25.6157 3.4219
9 1.1434 0.8746 0.10461 9.5593 0.11961 8.3605 32.6125 3.9008
10 1.1605 0.8617 0.09343 10.7027 0.10843 9.2222 40.3675 4.3772
11 1.1779 0.8489 0.08429 11.8633 0.09929 10.0711 48.8568 4.8512
12 1.1956 0.8364 0.07668 13.0412 0.09168 10.9075 58.0571 5.3227
13 1.2136 0.8240 0.07024 14.2368 0.08524 11.7315 67.9454 5.7917
14 1.2318 0.8118 0.06472 15.4504 0.07972 12.5434 78.4994 6.2582
15 1.2502 0.7999 0.05994 16.6821 0.07494 13.3432 89.6974 6.7223
16 1.2690 0.7880 0.05577 17.9324 0.07077 14.1313 101.5178 7.1839
17 1.2880 0.7764 0.05208 19.2014 0.06708 14.9076 113.9400 7.6431
18 1.3073 0.7649 0.04881 20.4894 0.06381 15.6726 126.9435 8.0997
19 1.3270 0.7536 0.04588 21.7967 0.06088 16.4262 140.5084 8.5539
20 1.3469 0.7425 0.04325 23.1237 0.05825 17.1686 154.6154 9.0057
21 1.3671 0.7315 0.04087 24.4705 0.05587 17.9001 169.2453 9.4550
22 1.3876 0.7207 0.03870 25.8376 0.05370 18.6208 184.3798 9.9018
23 1.4084 0.7100 0.03673 27.2251 0.05173 19.3309 200.0006 10.3462
24 1.4295 0.6995 0.03492 28.6335 0.04992 20.0304 216.0901 10.7881
25 1.4509 0.6892 0.03326 30.0630 0.04826 20.7196 232.6310 11.2276
26 1.4727 0.6790 0.03173 31.5140 0.04673 21.3986 249.6065 11.6646
27 1.4948 0.6690 0.03032 32.9867 0.04532 22.0676 267.0002 12.0992
28 1.5172 0.6591 0.02900 34.4815 0.04400 22.7267 284.7958 12.5313
29 1.5400 0.6494 0.02778 35.9987 0.04278 23.3761 302.9779 12.9610
30 1.5631 0.6398 0.02664 37.5387 0.04164 24.0158 321.5310 13.3883
36 1.7091 0.5851 0.02115 47.2760 0.03615 27.6607 439.8303 15.9009
40 1.8140 0.5513 0.01843 54.2679 0.03343 29.9158 524.3568 17.5277
48 2.0435 0.4894 0.01437 69.5652 0.02937 34.0426 703.5462 20.6667
50 2.1052 0.4750 0.01357 73.6828 0.02857 34.9997 749.9636 21.4277
52 2.1689 0.4611 0.01283 77.9249 0.02783 35.9287 796.8774 22.1794
55 2.2679 0.4409 0.01183 84.5296 0.02683 37.2715 868.0285 23.2894
60 2.4432 0.4093 0.01039 96.2147 0.02539 39.3803 988.1674 25.0930
72 2.9212 0.3423 0.00781 128.0772 0.02281 43.8447 1279.79 29.1893
75 3.0546 0.3274 0.00730 136.9728 0.02230 44.8416 1352.56 30.1631
84 3.4926 0.2863 0.00602 166.1726 0.02102 47.5786 1568.51 32.9668
90 3.8189 0.2619 0.00532 187.9299 0.02032 49.2099 1709.54 34.7399
96 4.1758 0.2395 0.00472 211.7202 0.01972 50.7017 1847.47 36.4381
100 4.4320 0.2256 0.00437 228.8030 0.01937 51.6247 1937.45 37.5295
108 4.9927 0.2003 0.00376 266.1778 0.01876 53.3137 2112.13 39.6171
120 5.9693 0.1675 0.00302 331.2882 0.01802 55.4985 2359.71 42.5185
132 7.1370 0.1401 0.00244 409.1354 0.01744 57.3257 2588.71 45.1579
144 8.5332 0.1172 0.00199 502.2109 0.01699 58.8540 2798.58 47.5512
240 35.6328 0.0281 0.00043 2308.85 0.01543 64.7957 3870.69 59.7368
360 212.7038 0.0047 0.00007 14114 0.01507 66.3532 4310.72 64.9662
480 1269.70 0.0008 0.00001 84580 0.01501 66.6142 4415.74 66.2883

Compound Interest Factor Tables 587
2% TABLE 7 Discrete Cash Flow: Compound Interest Factors 2%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.0200 0.9804 1.00000 1.0000 1.02000 0.9804
2 1.0404 0.9612 0.49505 2.0200 0.51505 1.9416 0.9612 0.4950
3 1.0612 0.9423 0.32675 3.0604 0.34675 2.8839 2.8458 0.9868
4 1.0824 0.9238 0.24262 4.1216 0.26262 3.8077 5.6173 1.4752
5 1.1041 0.9057 0.19216 5.2040 0.21216 4.7135 9.2403 1.9604
6 1.1262 0.8880 0.15853 6.3081 0.17853 5.6014 13.6801 2.4423
7 1.1487 0.8706 0.13451 7.4343 0.15451 6.4720 18.9035 2.9208
8 1.1717 0.8535 0.11651 8.5830 0.13651 7.3255 24.8779 3.3961
9 1.1951 0.8368 0.10252 9.7546 0.12252 8.1622 31.5720 3.8681
10 1.2190 0.8203 0.09133 10.9497 0.11133 8.9826 38.9551 4.3367
11 1.2434 0.8043 0.08218 12.1687 0.10218 9.7868 46.9977 4.8021
12 1.2682 0.7885 0.07456 13.4121 0.09456 10.5753 55.6712 5.2642
13 1.2936 0.7730 0.06812 14.6803 0.08812 11.3484 64.9475 5.7231
14 1.3195 0.7579 0.06260 15.9739 0.08260 12.1062 74.7999 6.1786
15 1.3459 0.7430 0.05783 17.2934 0.07783 12.8493 85.2021 6.6309
16 1.3728 0.7284 0.05365 18.6393 0.07365 13.5777 96.1288 7.0799
17 1.4002 0.7142 0.04997 20.0121 0.06997 14.2919 107.5554 7.5256
18 1.4282 0.7002 0.04670 21.4123 0.06670 14.9920 119.4581 7.9681
19 1.4568 0.6864 0.04378 22.8406 0.06378 15.6785 131.8139 8.4073
20 1.4859 0.6730 0.04116 24.2974 0.06116 16.3514 144.6003 8.8433
21 1.5157 0.6598 0.03878 25.7833 0.05878 17.0112 157.7959 9.2760
22 1.5460 0.6468 0.03663 27.2990 0.05663 17.6580 171.3795 9.7055
23 1.5769 0.6342 0.03467 28.8450 0.05467 18.2922 185.3309 10.1317
24 1.6084 0.6217 0.03287 30.4219 0.05287 18.9139 199.6305 10.5547
25 1.6406 0.6095 0.03122 32.0303 0.05122 19.5235 214.2592 10.9745
26 1.6734 0.5976 0.02970 33.6709 0.04970 20.1210 229.1987 11.3910
27 1.7069 0.5859 0.02829 35.3443 0.04829 20.7069 244.4311 11.8043
28 1.7410 0.5744 0.02699 37.0512 0.04699 21.2813 259.9392 12.2145
29 1.7758 0.5631 0.02578 38.7922 0.04578 21.8444 275.7064 12.6214
30 1.8114 0.5521 0.02465 40.5681 0.04465 22.3965 291.7164 13.0251
36 2.0399 0.4902 0.01923 51.9944 0.03923 25.4888 392.0405 15.3809
40 2.2080 0.4529 0.01656 60.4020 0.03656 27.3555 461.9931 16.8885
48 2.5871 0.3865 0.01260 79.3535 0.03260 30.6731 605.9657 19.7556
50 2.6916 0.3715 0.01182 84.5794 0.03182 31.4236 642.3606 20.4420
52 2.8003 0.3571 0.01111 90.0164 0.03111 32.1449 678.7849 21.1164
55 2.9717 0.3365 0.01014 98.5865 0.03014 33.1748 733.3527 22.1057
60 3.2810 0.3048 0.00877 114.0515 0.02877 34.7609 823.6975 23.6961
72 4.1611 0.2403 0.00633 158.0570 0.02633 37.9841 1034.06 27.2234
75 4.4158 0.2265 0.00586 170.7918 0.02586 38.6771 1084.64 28.0434
84 5.2773 0.1895 0.00468 213.8666 0.02468 40.5255 1230.42 30.3616
90 5.9431 0.1683 0.00405 247.1567 0.02405 41.5869 1322.17 31.7929
96 6.6929 0.1494 0.00351 284.6467 0.02351 42.5294 1409.30 33.1370
100 7.2446 0.1380 0.00320 312.2323 0.02320 43.0984 1464.75 33.9863
108 8.4883 0.1178 0.00267 374.4129 0.02267 44.1095 1569.30 35.5774
120 10.7652 0.0929 0.00205 488.2582 0.02205 45.3554 1710.42 37.7114
132 13.6528 0.0732 0.00158 632.6415 0.02158 46.3378 1833.47 39.5676
144 17.3151 0.0578 0.00123 815.7545 0.02123 47.1123 1939.79 41.1738
240 115.8887 0.0086 0.00017 5744.44 0.02017 49.5686 2374.88 47.9110
360 1247.56 0.0008 0.00002 62328 0.02002 49.9599 2482.57 49.7112
480 13430 0.0001 0.02000 49.9963 2498.03 49.9643

588 Compound Interest Factor Tables
3% TABLE 8 Discrete Cash Flow: Compound Interest Factors 3%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.0300 0.9709 1.00000 1.0000 1.03000 0.9709
2 1.0609 0.9426 0.49261 2.0300 0.52261 1.9135 0.9426 0.4926
3 1.0927 0.9151 0.32353 3.0909 0.35353 2.8286 2.7729 0.9803
4 1.1255 0.8885 0.23903 4.1836 0.26903 3.7171 5.4383 1.4631
5 1.1593 0.8626 0.18835 5.3091 0.21835 4.5797 8.8888 1.9409
6 1.1941 0.8375 0.15460 6.4684 0.18460 5.4172 13.0762 2.4138
7 1.2299 0.8131 0.13051 7.6625 0.16051 6.2303 17.9547 2.8819
8 1.2668 0.7894 0.11246 8.8923 0.14246 7.0197 23.4806 3.3450
9 1.3048 0.7664 0.09843 10.1591 0.12843 7.7861 29.6119 3.8032
10 1.3439 0.7441 0.08723 11.4639 0.11723 8.5302 36.3088 4.2565
11 1.3842 0.7224 0.07808 12.8078 0.10808 9.2526 43.5330 4.7049
12 1.4258 0.7014 0.07046 14.1920 0.10046 9.9540 51.2482 5.1485
13 1.4685 0.6810 0.06403 15.6178 0.09403 10.6350 59.4196 5.5872
14 1.5126 0.6611 0.05853 17.0863 0.08853 11.2961 68.0141 6.0210
15 1.5580 0.6419 0.05377 18.5989 0.08377 11.9379 77.0002 6.4500
16 1.6047 0.6232 0.04961 20.1569 0.07961 12.5611 86.3477 6.8742
17 1.6528 0.6050 0.04595 21.7616 0.07595 13.1661 96.0280 7.2936
18 1.7024 0.5874 0.04271 23.4144 0.07271 13.7535 106.0137 7.7081
19 1.7535 0.5703 0.03981 25.1169 0.06981 14.3238 116.2788 8.1179
20 1.8061 0.5537 0.03722 26.8704 0.06722 14.8775 126.7987 8.5229
21 1.8603 0.5375 0.03487 28.6765 0.06487 15.4150 137.5496 8.9231
22 1.9161 0.5219 0.03275 30.5368 0.06275 15.9369 148.5094 9.3186
23 1.9736 0.5067 0.03081 32.4529 0.06081 16.4436 159.6566 9.7093
24 2.0328 0.4919 0.02905 34.4265 0.05905 16.9355 170.9711 10.0954
25 2.0938 0.4776 0.02743 36.4593 0.05743 17.4131 182.4336 10.4768
26 2.1566 0.4637 0.02594 38.5530 0.05594 17.8768 194.0260 10.8535
27 2.2213 0.4502 0.02456 40.7096 0.05456 18.3270 205.7309 11.2255
28 2.2879 0.4371 0.02329 42.9309 0.05329 18.7641 217.5320 11.5930
29 2.3566 0.4243 0.02211 45.2189 0.05211 19.1885 229.4137 11.9558
30 2.4273 0.4120 0.02102 47.5754 0.05102 19.6004 241.3613 12.3141
31 2.5001 0.4000 0.02000 50.0027 0.05000 20.0004 253.3609 12.6678
32 2.5751 0.3883 0.01905 52.5028 0.04905 20.3888 265.3993 13.0169
33 2.6523 0.3770 0.01816 55.0778 0.04816 20.7658 277.4642 13.3616
34 2.7319 0.3660 0.01732 57.7302 0.04732 21.1318 289.5437 13.7018
35 2.8139 0.3554 0.01654 60.4621 0.04654 21.4872 301.6267 14.0375
40 3.2620 0.3066 0.01326 75.4013 0.04326 23.1148 361.7499 15.6502
45 3.7816 0.2644 0.01079 92.7199 0.04079 24.5187 420.6325 17.1556
50 4.3839 0.2281 0.00887 112.7969 0.03887 25.7298 477.4803 18.5575
55 5.0821 0.1968 0.00735 136.0716 0.03735 26.7744 531.7411 19.8600
60 5.8916 0.1697 0.00613 163.0534 0.03613 27.6756 583.0526 21.0674
65 6.8300 0.1464 0.00515 194.3328 0.03515 28.4529 631.2010 22.1841
70 7.9178 0.1263 0.00434 230.5941 0.03434 29.1234 676.0869 23.2145
75 9.1789 0.1089 0.00367 272.6309 0.03367 29.7018 717.6978 24.1634
80 10.6409 0.0940 0.00311 321.3630 0.03311 30.2008 756.0865 25.0353
84 11.9764 0.0835 0.00273 365.8805 0.03273 30.5501 784.5434 25.6806
85 12.3357 0.0811 0.00265 377.8570 0.03265 30.6312 791.3529 25.8349
90 14.3005 0.0699 0.00226 443.3489 0.03226 31.0024 823.6302 26.5667
96 17.0755 0.0586 0.00187 535.8502 0.03187 31.3812 858.6377 27.3615
108 24.3456 0.0411 0.00129 778.1863 0.03129 31.9642 917.6013 28.7072
120 34.7110 0.0288 0.00089 1123.70 0.03089 32.3730 963.8635 29.7737

Compound Interest Factor Tables 589
4% TABLE 9 Discrete Cash Flow: Compound Interest Factors 4%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.0400 0.9615 1.00000 1.0000 1.04000 0.9615
2 1.0816 0.9246 0.49020 2.0400 0.53020 1.8861 0.9246 0.4902
3 1.1249 0.8890 0.32035 3.1216 0.36035 2.7751 2.7025 0.9739
4 1.1699 0.8548 0.23549 4.2465 0.27549 3.6299 5.2670 1.4510
5 1.2167 0.8219 0.18463 5.4163 0.22463 4.4518 8.5547 1.9216
6 1.2653 0.7903 0.15076 6.6330 0.19076 5.2421 12.5062 2.3857
7 1.3159 0.7599 0.12661 7.8983 0.16661 6.0021 17.0657 2.8433
8 1.3686 0.7307 0.10853 9.2142 0.14853 6.7327 22.1806 3.2944
9 1.4233 0.7026 0.09449 10.5828 0.13449 7.4353 27.8013 3.7391
10 1.4802 0.6756 0.08329 12.0061 0.12329 8.1109 33.8814 4.1773
11 1.5395 0.6496 0.07415 13.4864 0.11415 8.7605 40.3772 4.6090
12 1.6010 0.6246 0.06655 15.0258 0.10655 9.3851 47.2477 5.0343
13 1.6651 0.6006 0.06014 16.6268 0.10014 9.9856 54.4546 5.4533
14 1.7317 0.5775 0.05467 18.2919 0.09467 10.5631 61.9618 5.8659
15 1.8009 0.5553 0.04994 20.0236 0.08994 11.1184 69.7355 6.2721
16 1.8730 0.5339 0.04582 21.8245 0.08582 11.6523 77.7441 6.6720
17 1.9479 0.5134 0.04220 23.6975 0.08220 12.1657 85.9581 7.0656
18 2.0258 0.4936 0.03899 25.6454 0.07899 12.6593 94.3498 7.4530
19 2.1068 0.4746 0.03614 27.6712 0.07614 13.1339 102.8933 7.8342
20 2.1911 0.4564 0.03358 29.7781 0.07358 13.5903 111.5647 8.2091
21 2.2788 0.4388 0.03128 31.9692 0.07128 14.0292 120.3414 8.5779
22 2.3699 0.4220 0.02920 34.2480 0.06920 14.4511 129.2024 8.9407
23 2.4647 0.4057 0.02731 36.6179 0.06731 14.8568 138.1284 9.2973
24 2.5633 0.3901 0.02559 39.0826 0.06559 15.2470 147.1012 9.6479
25 2.6658 0.3751 0.02401 41.6459 0.06401 15.6221 156.1040 9.9925
26 2.7725 0.3607 0.02257 44.3117 0.06257 15.9828 165.1212 10.3312
27 2.8834 0.3468 0.02124 47.0842 0.06124 16.3296 174.1385 10.6640
28 2.9987 0.3335 0.02001 49.9676 0.06001 16.6631 183.1424 10.9909
29 3.1187 0.3207 0.01888 52.9663 0.05888 16.9837 192.1206 11.3120
30 3.2434 0.3083 0.01783 56.0849 0.05783 17.2920 201.0618 11.6274
31 3.3731 0.2965 0.01686 59.3283 0.05686 17.5885 209.9556 11.9371
32 3.5081 0.2851 0.01595 62.7015 0.05595 17.8736 218.7924 12.2411
33 3.6484 0.2741 0.01510 66.2095 0.05510 18.1476 227.5634 12.5396
34 3.7943 0.2636 0.01431 69.8579 0.05431 18.4112 236.2607 12.8324
35 3.9461 0.2534 0.01358 73.6522 0.05358 18.6646 244.8768 13.1198
40 4.8010 0.2083 0.01052 95.0255 0.05052 19.7928 286.5303 14.4765
45 5.8412 0.1712 0.00826 121.0294 0.04826 20.7200 325.4028 15.7047
50 7.1067 0.1407 0.00655 152.6671 0.04655 21.4822 361.1638 16.8122
55 8.6464 0.1157 0.00523 191.1592 0.04523 22.1086 393.6890 17.8070
60 10.5196 0.0951 0.00420 237.9907 0.04420 22.6235 422.9966 18.6972
65 12.7987 0.0781 0.00339 294.9684 0.04339 23.0467 449.2014 19.4909
70 15.5716 0.0642 0.00275 364.2905 0.04275 23.3945 472.4789 20.1961
75 18.9453 0.0528 0.00223 448.6314 0.04223 23.6804 493.0408 20.8206
80 23.0498 0.0434 0.00181 551.2450 0.04181 23.9154 511.1161 21.3718
85 28.0436 0.0357 0.00148 676.0901 0.04148 24.1085 526.9384 21.8569
90 34.1193 0.0293 0.00121 827.9833 0.04121 24.2673 540.7369 22.2826
96 43.1718 0.0232 0.00095 1054.30 0.04095 24.4209 554.9312 22.7236
108 69.1195 0.0145 0.00059 1702.99 0.04059 24.6383 576.8949 23.4146
120 110.6626 0.0090 0.00036 2741.56 0.04036 24.7741 592.2428 23.9057
144 283.6618 0.0035 0.00014 7066.55 0.04014 24.9119 610.1055 24.4906

590 Compound Interest Factor Tables
5% TABLE 10 Discrete Cash Flow: Compound Interest Factors 5%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.0500 0.9524 1.00000 1.0000 1.05000 0.9524
2 1.1025 0.9070 0.48780 2.0500 0.53780 1.8594 0.9070 0.4878
3 1.1576 0.8638 0.31721 3.1525 0.36721 2.7232 2.6347 0.9675
4 1.2155 0.8227 0.23201 4.3101 0.28201 3.5460 5.1028 1.4391
5 1.2763 0.7835 0.18097 5.5256 0.23097 4.3295 8.2369 1.9025
6 1.3401 0.7462 0.14702 6.8019 0.19702 5.0757 11.9680 2.3579
7 1.4071 0.7107 0.12282 8.1420 0.17282 5.7864 16.2321 2.8052
8 1.4775 0.6768 0.10472 9.5491 0.15472 6.4632 20.9700 3.2445
9 1.5513 0.6446 0.09069 11.0266 0.14069 7.1078 26.1268 3.6758
10 1.6289 0.6139 0.07950 12.5779 0.12950 7.7217 31.6520 4.0991
11 1.7103 0.5847 0.07039 14.2068 0.12039 8.3064 37.4988 4.5144
12 1.7959 0.5568 0.06283 15.9171 0.11283 8.8633 43.6241 4.9219
13 1.8856 0.5303 0.05646 17.7130 0.10646 9.3936 49.9879 5.3215
14 1.9799 0.5051 0.05102 19.5986 0.10102 9.8986 56.5538 5.7133
15 2.0789 0.4810 0.04634 21.5786 0.09634 10.3797 63.2880 6.0973
16 2.1829 0.4581 0.04227 23.6575 0.09227 10.8378 70.1597 6.4736
17 2.2920 0.4363 0.03870 25.8404 0.08870 11.2741 77.1405 6.8423
18 2.4066 0.4155 0.03555 28.1324 0.08555 11.6896 84.2043 7.2034
19 2.5270 0.3957 0.03275 30.5390 0.08275 12.0853 91.3275 7.5569
20 2.6533 0.3769 0.03024 33.0660 0.08024 12.4622 98.4884 7.9030
21 2.7860 0.3589 0.02800 35.7193 0.07800 12.8212 105.6673 8.2416
22 2.9253 0.3418 0.02597 38.5052 0.07597 13.1630 112.8461 8.5730
23 3.0715 0.3256 0.02414 41.4305 0.07414 13.4886 120.0087 8.8971
24 3.2251 0.3101 0.02247 44.5020 0.07247 13.7986 127.1402 9.2140
25 3.3864 0.2953 0.02095 47.7271 0.07095 14.0939 134.2275 9.5238
26 3.5557 0.2812 0.01956 51.1135 0.06956 14.3752 141.2585 9.8266
27 3.7335 0.2678 0.01829 54.6691 0.06829 14.6430 148.2226 10.1224
28 3.9201 0.2551 0.01712 58.4026 0.06712 14.8981 155.1101 10.4114
29 4.1161 0.2429 0.01605 62.3227 0.06605 15.1411 161.9126 10.6936
30 4.3219 0.2314 0.01505 66.4388 0.06505 15.3725 168.6226 10.9691
31 4.5380 0.2204 0.01413 70.7608 0.06413 15.5928 175.2333 11.2381
32 4.7649 0.2099 0.01328 75.2988 0.06328 15.8027 181.7392 11.5005
33 5.0032 0.1999 0.01249 80.0638 0.06249 16.0025 188.1351 11.7566
34 5.2533 0.1904 0.01176 85.0670 0.06176 16.1929 194.4168 12.0063
35 5.5160 0.1813 0.01107 90.3203 0.06107 16.3742 200.5807 12.2498
40 7.0400 0.1420 0.00828 120.7998 0.05828 17.1591 229.5452 13.3775
45 8.9850 0.1113 0.00626 159.7002 0.05626 17.7741 255.3145 14.3644
50 11.4674 0.0872 0.00478 209.3480 0.05478 18.2559 277.9148 15.2233
55 14.6356 0.0683 0.00367 272.7126 0.05367 18.6335 297.5104 15.9664
60 18.6792 0.0535 0.00283 353.5837 0.05283 18.9293 314.3432 16.6062
65 23.8399 0.0419 0.00219 456.7980 0.05219 19.1611 328.6910 17.1541
70 30.4264 0.0329 0.00170 588.5285 0.05170 19.3427 340.8409 17.6212
75 38.8327 0.0258 0.00132 756.6537 0.05132 19.4850 351.0721 18.0176
80 49.5614 0.0202 0.00103 971.2288 0.05103 19.5965 359.6460 18.3526
85 63.2544 0.0158 0.00080 1245.09 0.05080 19.6838 366.8007 18.6346
90 80.7304 0.0124 0.00063 1594.61 0.05063 19.7523 372.7488 18.8712
95 103.0347 0.0097 0.00049 2040.69 0.05049 19.8059 377.6774 19.0689
96 108.1864 0.0092 0.00047 2143.73 0.05047 19.8151 378.5555 19.1044
98 119.2755 0.0084 0.00042 2365.51 0.05042 19.8323 380.2139 19.1714
100 131.5013 0.0076 0.00038 2610.03 0.05038 19.8479 381.7492 19.2337

Compound Interest Factor Tables 591
6% TABLE 11 Discrete Cash Flow: Compound Interest Factors 6%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.0600 0.9434 1.00000 1.0000 1.06000 0.9434
2 1.1236 0.8900 0.48544 2.0600 0.54544 1.8334 0.8900 0.4854
3 1.1910 0.8396 0.31411 3.1836 0.37411 2.6730 2.5692 0.9612
4 1.2625 0.7921 0.22859 4.3746 0.28859 3.4651 4.9455 1.4272
5 1.3382 0.7473 0.17740 5.6371 0.23740 4.2124 7.9345 1.8836
6 1.4185 0.7050 0.14336 6.9753 0.20336 4.9173 11.4594 2.3304
7 1.5036 0.6651 0.11914 8.3938 0.17914 5.5824 15.4497 2.7676
8 1.5938 0.6274 0.10104 9.8975 0.16104 6.2098 19.8416 3.1952
9 1.6895 0.5919 0.08702 11.4913 0.14702 6.8017 24.5768 3.6133
10 1.7908 0.5584 0.07587 13.1808 0.13587 7.3601 29.6023 4.0220
11 1.8983 0.5268 0.06679 14.9716 0.12679 7.8869 34.8702 4.4213
12 2.0122 0.4970 0.05928 16.8699 0.11928 8.3838 40.3369 4.8113
13 2.1329 0.4688 0.05296 18.8821 0.11296 8.8527 45.9629 5.1920
14 2.2609 0.4423 0.04758 21.0151 0.10758 9.2950 51.7128 5.5635
15 2.3966 0.4173 0.04296 23.2760 0.10296 9.7122 57.5546 5.9260
16 2.5404 0.3936 0.03895 25.6725 0.09895 10.1059 63.4592 6.2794
17 2.6928 0.3714 0.03544 28.2129 0.09544 10.4773 69.4011 6.6240
18 2.8543 0.3503 0.03236 30.9057 0.09236 10.8276 75.3569 6.9597
19 3.0256 0.3305 0.02962 33.7600 0.08962 11.1581 81.3062 7.2867
20 3.2071 0.3118 0.02718 36.7856 0.08718 11.4699 87.2304 7.6051
21 3.3996 0.2942 0.02500 39.9927 0.08500 11.7641 93.1136 7.9151
22 3.6035 0.2775 0.02305 43.3923 0.08305 12.0416 98.9412 8.2166
23 3.8197 0.2618 0.02128 46.9958 0.08128 12.3034 104.7007 8.5099
24 4.0489 0.2470 0.01968 50.8156 0.07968 12.5504 110.3812 8.7951
25 4.2919 0.2330 0.01823 54.8645 0.07823 12.7834 115.9732 9.0722
26 4.5494 0.2198 0.01690 59.1564 0.07690 13.0032 121.4684 9.3414
27 4.8223 0.2074 0.01570 63.7058 0.07570 13.2105 126.8600 9.6029
28 5.1117 0.1956 0.01459 68.5281 0.07459 13.4062 132.1420 9.8568
29 5.4184 0.1846 0.01358 73.6398 0.07358 13.5907 137.3096 10.1032
30 5.7435 0.1741 0.01265 79.0582 0.07265 13.7648 142.3588 10.3422
31 6.0881 0.1643 0.01179 84.8017 0.07179 13.9291 147.2864 10.5740
32 6.4534 0.1550 0.01100 90.8898 0.07100 14.0840 152.0901 10.7988
33 6.8406 0.1462 0.01027 97.3432 0.07027 14.2302 156.7681 11.0166
34 7.2510 0.1379 0.00960 104.1838 0.06960 14.3681 161.3192 11.2276
35 7.6861 0.1301 0.00897 111.4348 0.06897 14.4982 165.7427 11.4319
40 10.2857 0.0972 0.00646 154.7620 0.06646 15.0463 185.9568 12.3590
45 13.7646 0.0727 0.00470 212.7435 0.06470 15.4558 203.1096 13.1413
50 18.4202 0.0543 0.00344 290.3359 0.06344 15.7619 217.4574 13.7964
55 24.6503 0.0406 0.00254 394.1720 0.06254 15.9905 229.3222 14.3411
60 32.9877 0.0303 0.00188 533.1282 0.06188 16.1614 239.0428 14.7909
65 44.1450 0.0227 0.00139 719.0829 0.06139 16.2891 246.9450 15.1601
70 59.0759 0.0169 0.00103 967.9322 0.06103 16.3845 253.3271 15.4613
75 79.0569 0.0126 0.00077 1300.95 0.06077 16.4558 258.4527 15.7058
80 105.7960 0.0095 0.00057 1746.60 0.06057 16.5091 262.5493 15.9033
85 141.5789 0.0071 0.00043 2342.98 0.06043 16.5489 265.8096 16.0620
90 189.4645 0.0053 0.00032 3141.08 0.06032 16.5787 268.3946 16.1891
95 253.5463 0.0039 0.00024 4209.10 0.06024 16.6009 270.4375 16.2905
96 268.7590 0.0037 0.00022 4462.65 0.06022 16.6047 270.7909 16.3081
98 301.9776 0.0033 0.00020 5016.29 0.06020 16.6115 271.4491 16.3411
100 339.3021 0.0029 0.00018 5638.37 0.06018 16.6175 272.0471 16.3711

592 Compound Interest Factor Tables
7% TABLE 12 Discrete Cash Flow: Compound Interest Factors 7%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.0700 0.9346 1.00000 1.0000 1.07000 0.9346
2 1.1449 0.8734 0.48309 2.0700 0.55309 1.8080 0.8734 0.4831
3 1.2250 0.8163 0.31105 3.2149 0.38105 2.6243 2.5060 0.9549
4 1.3108 0.7629 0.22523 4.4399 0.29523 3.3872 4.7947 1.4155
5 1.4026 0.7130 0.17389 5.7507 0.24389 4.1002 7.6467 1.8650
6 1.5007 0.6663 0.13980 7.1533 0.20980 4.7665 10.9784 2.3032
7 1.6058 0.6227 0.11555 8.6540 0.18555 5.3893 14.7149 2.7304
8 1.7182 0.5820 0.09747 10.2598 0.16747 5.9713 18.7889 3.1465
9 1.8385 0.5439 0.08349 11.9780 0.15349 6.5152 23.1404 3.5517
10 1.9672 0.5083 0.07238 13.8164 0.14238 7.0236 27.7156 3.9461
11 2.1049 0.4751 0.06336 15.7836 0.13336 7.4987 32.4665 4.3296
12 2.2522 0.4440 0.05590 17.8885 0.12590 7.9427 37.3506 4.7025
13 2.4098 0.4150 0.04965 20.1406 0.11965 8.3577 42.3302 5.0648
14 2.5785 0.3878 0.04434 22.5505 0.11434 8.7455 47.3718 5.4167
15 2.7590 0.3624 0.03979 25.1290 0.10979 9.1079 52.4461 5.7583
16 2.9522 0.3387 0.03586 27.8881 0.10586 9.4466 57.5271 6.0897
17 3.1588 0.3166 0.03243 30.8402 0.10243 9.7632 62.5923 6.4110
18 3.3799 0.2959 0.02941 33.9990 0.09941 10.0591 67.6219 6.7225
19 3.6165 0.2765 0.02675 37.3790 0.09675 10.3356 72.5991 7.0242
20 3.8697 0.2584 0.02439 40.9955 0.09439 10.5940 77.5091 7.3163
21 4.1406 0.2415 0.02229 44.8652 0.09229 10.8355 82.3393 7.5990
22 4.4304 0.2257 0.02041 49.0057 0.09041 11.0612 87.0793 7.8725
23 4.7405 0.2109 0.01871 53.4361 0.08871 11.2722 91.7201 8.1369
24 5.0724 0.1971 0.01719 58.1767 0.08719 11.4693 96.2545 8.3923
25 5.4274 0.1842 0.01581 63.2490 0.08581 11.6536 100.6765 8.6391
26 5.8074 0.1722 0.01456 68.6765 0.08456 11.8258 104.9814 8.8773
27 6.2139 0.1609 0.01343 74.4838 0.08343 11.9867 109.1656 9.1072
28 6.6488 0.1504 0.01239 80.6977 0.08239 12.1371 113.2264 9.3289
29 7.1143 0.1406 0.01145 87.3465 0.08145 12.2777 117.1622 9.5427
30 7.6123 0.1314 0.01059 94.4608 0.08059 12.4090 120.9718 9.7487
31 8.1451 0.1228 0.00980 102.0730 0.07980 12.5318 124.6550 9.9471
32 8.7153 0.1147 0.00907 110.2182 0.07907 12.6466 128.2120 10.1381
33 9.3253 0.1072 0.00841 118.9334 0.07841 12.7538 131.6435 10.3219
34 9.9781 0.1002 0.00780 128.2588 0.07780 12.8540 134.9507 10.4987
35 10.6766 0.0937 0.00723 138.2369 0.07723 12.9477 138.1353 10.6687
40 14.9745 0.0668 0.00501 199.6351 0.07501 13.3317 152.2928 11.4233
45 21.0025 0.0476 0.00350 285.7493 0.07350 13.6055 163.7559 12.0360
50 29.4570 0.0339 0.00246 406.5289 0.07246 13.8007 172.9051 12.5287
55 41.3150 0.0242 0.00174 575.9286 0.07174 13.9399 180.1243 12.9215
60 57.9464 0.0173 0.00123 813.5204 0.07123 14.0392 185.7677 13.2321
65 81.2729 0.0123 0.00087 1146.76 0.07087 14.1099 190.1452 13.4760
70 113.9894 0.0088 0.00062 1614.13 0.07062 14.1604 193.5185 13.6662
75 159.8760 0.0063 0.00044 2269.66 0.07044 14.1964 196.1035 13.8136
80 224.2344 0.0045 0.00031 3189.06 0.07031 14.2220 198.0748 13.9273
85 314.5003 0.0032 0.00022 4478.58 0.07022 14.2403 199.5717 14.0146
90 441.1030 0.0023 0.00016 6287.19 0.07016 14.2533 200.7042 14.0812
95 618.6697 0.0016 0.00011 8823.85 0.07011 14.2626 201.5581 14.1319
96 661.9766 0.0015 0.00011 9442.52 0.07011 14.2641 201.7016 14.1405
98 757.8970 0.0013 0.00009 10813 0.07009 14.2669 201.9651 14.1562
100 867.7163 0.0012 0.00008 12382 0.07008 14.2693 202.2001 14.1703

Compound Interest Factor Tables 593
8% TABLE 13 Discrete Cash Flow: Compound Interest Factors 8%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.0800 0.9259 1.00000 1.0000 1.08000 0.9259
2 1.1664 0.8573 0.48077 2.0800 0.56077 1.7833 0.8573 0.4808
3 1.2597 0.7938 0.30803 3.2464 0.38803 2.5771 2.4450 0.9487
4 1.3605 0.7350 0.22192 4.5061 0.30192 3.3121 4.6501 1.4040
5 1.4693 0.6806 0.17046 5.8666 0.25046 3.9927 7.3724 1.8465
6 1.5869 0.6302 0.13632 7.3359 0.21632 4.6229 10.5233 2.2763
7 1.7138 0.5835 0.11207 8.9228 0.19207 5.2064 14.0242 2.6937
8 1.8509 0.5403 0.09401 10.6366 0.17401 5.7466 17.8061 3.0985
9 1.9990 0.5002 0.08008 12.4876 0.16008 6.2469 21.8081 3.4910
10 2.1589 0.4632 0.06903 14.4866 0.14903 6.7101 25.9768 3.8713
11 2.3316 0.4289 0.06008 16.6455 0.14008 7.1390 30.2657 4.2395
12 2.5182 0.3971 0.05270 18.9771 0.13270 7.5361 34.6339 4.5957
13 2.7196 0.3677 0.04652 21.4953 0.12652 7.9038 39.0463 4.9402
14 2.9372 0.3405 0.04130 24.2149 0.12130 8.2442 43.4723 5.2731
15 3.1722 0.3152 0.03683 27.1521 0.11683 8.5595 47.8857 5.5945
16 3.4259 0.2919 0.03298 30.3243 0.11298 8.8514 52.2640 5.9046
17 3.7000 0.2703 0.02963 33.7502 0.10963 9.1216 56.5883 6.2037
18 3.9960 0.2502 0.02670 37.4502 0.10670 9.3719 60.8426 6.4920
19 4.3157 0.2317 0.02413 41.4463 0.10413 9.6036 65.0134 6.7697
20 4.6610 0.2145 0.02185 45.7620 0.10185 9.8181 69.0898 7.0369
21 5.0338 0.1987 0.01983 50.4229 0.09983 10.0168 73.0629 7.2940
22 5.4365 0.1839 0.01803 55.4568 0.09803 10.2007 76.9257 7.5412
23 5.8715 0.1703 0.01642 60.8933 0.09642 10.3711 80.6726 7.7786
24 6.3412 0.1577 0.01498 66.7648 0.09498 10.5288 84.2997 8.0066
25 6.8485 0.1460 0.01368 73.1059 0.09368 10.6748 87.8041 8.2254
26 7.3964 0.1352 0.01251 79.9544 0.09251 10.8100 91.1842 8.4352
27 7.9881 0.1252 0.01145 87.3508 0.09145 10.9352 94.4390 8.6363
28 8.6271 0.1159 0.01049 95.3388 0.09049 11.0511 97.5687 8.8289
29 9.3173 0.1073 0.00962 103.9659 0.08962 11.1584 100.5738 9.0133
30 10.0627 0.0994 0.00883 113.2832 0.08883 11.2578 103.4558 9.1897
31 10.8677 0.0920 0.00811 123.3459 0.08811 11.3498 106.2163 9.3584
32 11.7371 0.0852 0.00745 134.2135 0.08745 11.4350 108.8575 9.5197
33 12.6760 0.0789 0.00685 145.9506 0.08685 11.5139 111.3819 9.6737
34 13.6901 0.0730 0.00630 158.6267 0.08630 11.5869 113.7924 9.8208
35 14.7853 0.0676 0.00580 172.3168 0.08580 11.6546 116.0920 9.9611
40 21.7245 0.0460 0.00386 259.0565 0.08386 11.9246 126.0422 10.5699
45 31.9204 0.0313 0.00259 386.5056 0.08259 12.1084 133.7331 11.0447
50 46.9016 0.0213 0.00174 573.7702 0.08174 12.2335 139.5928 11.4107
55 68.9139 0.0145 0.00118 848.9232 0.08118 12.3186 144.0065 11.6902
60 101.2571 0.0099 0.00080 1253.21 0.08080 12.3766 147.3000 11.9015
65 148.7798 0.0067 0.00054 1847.25 0.08054 12.4160 149.7387 12.0602
70 218.6064 0.0046 0.00037 2720.08 0.08037 12.4428 151.5326 12.1783
75 321.2045 0.0031 0.00025 4002.56 0.08025 12.4611 152.8448 12.2658
80 471.9548 0.0021 0.00017 5886.94 0.08017 12.4735 153.8001 12.3301
85 693.4565 0.0014 0.00012 8655.71 0.08012 12.4820 154.4925 12.3772
90 1018.92 0.0010 0.00008 12724 0.08008 12.4877 154.9925 12.4116
95 1497.12 0.0007 0.00005 18702 0.08005 12.4917 155.3524 12.4365
96 1616.89 0.0006 0.00005 20199 0.08005 12.4923 155.4112 12.4406
98 1885.94 0.0005 0.00004 23562 0.08004 12.4934 155.5176 12.4480
100 2199.76 0.0005 0.00004 27485 0.08004 12.4943 155.6107 12.4545

594 Compound Interest Factor Tables
9% TABLE 14 Discrete Cash Flow: Compound Interest Factors 9%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.0900 0.9174 1.00000 1.0000 1.09000 0.9174
2 1.1881 0.8417 0.47847 2.0900 0.56847 1.7591 0.8417 0.4785
3 1.2950 0.7722 0.30505 3.2781 0.39505 2.5313 2.3860 0.9426
4 1.4116 0.7084 0.21867 4.5731 0.30867 3.2397 4.5113 1.3925
5 1.5386 0.6499 0.16709 5.9847 0.25709 3.8897 7.1110 1.8282
6 1.6771 0.5963 0.13292 7.5233 0.22292 4.4859 10.0924 2.2498
7 1.8280 0.5470 0.10869 9.2004 0.19869 5.0330 13.3746 2.6574
8 1.9926 0.5019 0.09067 11.0285 0.18067 5.5348 16.8877 3.0512
9 2.1719 0.4604 0.07680 13.0210 0.16680 5.9952 20.5711 3.4312
10 2.3674 0.4224 0.06582 15.1929 0.15582 6.4177 24.3728 3.7978
11 2.5804 0.3875 0.05695 17.5603 0.14695 6.8052 28.2481 4.1510
12 2.8127 0.3555 0.04965 20.1407 0.13965 7.1607 32.1590 4.4910
13 3.0658 0.3262 0.04357 22.9534 0.13357 7.4869 36.0731 4.8182
14 3.3417 0.2992 0.03843 26.0192 0.12843 7.7862 39.9633 5.1326
15 3.6425 0.2745 0.03406 29.3609 0.12406 8.0607 43.8069 5.4346
16 3.9703 0.2519 0.03030 33.0034 0.12030 8.3126 47.5849 5.7245
17 4.3276 0.2311 0.02705 36.9737 0.11705 8.5436 51.2821 6.0024
18 4.7171 0.2120 0.02421 41.3013 0.11421 8.7556 54.8860 6.2687
19 5.1417 0.1945 0.02173 46.0185 0.11173 8.9501 58.3868 6.5236
20 5.6044 0.1784 0.01955 51.1601 0.10955 9.1285 61.7770 6.7674
21 6.1088 0.1637 0.01762 56.7645 0.10762 9.2922 65.0509 7.0006
22 6.6586 0.1502 0.01590 62.8733 0.10590 9.4424 68.2048 7.2232
23 7.2579 0.1378 0.01438 69.5319 0.10438 9.5802 71.2359 7.4357
24 7.9111 0.1264 0.01302 76.7898 0.10302 9.7066 74.1433 7.6384
25 8.6231 0.1160 0.01181 84.7009 0.10181 9.8226 76.9265 7.8316
26 9.3992 0.1064 0.01072 93.3240 0.10072 9.9290 79.5863 8.0156
27 10.2451 0.0976 0.00973 102.7231 0.09973 10.0266 82.1241 8.1906
28 11.1671 0.0895 0.00885 112.9682 0.09885 10.1161 84.5419 8.3571
29 12.1722 0.0822 0.00806 124.1354 0.09806 10.1983 86.8422 8.5154
30 13.2677 0.0754 0.00734 136.3075 0.09734 10.2737 89.0280 8.6657
31 14.4618 0.0691 0.00669 149.5752 0.09669 10.3428 91.1024 8.8083
32 15.7633 0.0634 0.00610 164.0370 0.09610 10.4062 93.0690 8.9436
33 17.1820 0.0582 0.00556 179.8003 0.09556 10.4644 94.9314 9.0718
34 18.7284 0.0534 0.00508 196.9823 0.09508 10.5178 96.6935 9.1933
35 20.4140 0.0490 0.00464 215.7108 0.09464 10.5668 98.3590 9.3083
40 31.4094 0.0318 0.00296 337.8824 0.09296 10.7574 105.3762 9.7957
45 48.3273 0.0207 0.00190 525.8587 0.09190 10.8812 110.5561 10.1603
50 74.3575 0.0134 0.00123 815.0836 0.09123 10.9617 114.3251 10.4295
55 114.4083 0.0087 0.00079 1260.09 0.09079 11.0140 117.0362 10.6261
60 176.0313 0.0057 0.00051 1944.79 0.09051 11.0480 118.9683 10.7683
65 270.8460 0.0037 0.00033 2998.29 0.09033 11.0701 120.3344 10.8702
70 416.7301 0.0024 0.00022 4619.22 0.09022 11.0844 121.2942 10.9427
75 641.1909 0.0016 0.00014 7113.23 0.09014 11.0938 121.9646 10.9940
80 986.5517 0.0010 0.00009 10951 0.09009 11.0998 122.4306 11.0299
85 1517.93 0.0007 0.00006 16855 0.09006 11.1038 122.7533 11.0551
90 2335.53 0.0004 0.00004 25939 0.09004 11.1064 122.9758 11.0726
95 3593.50 0.0003 0.00003 39917 0.09003 11.1080 123.1287 11.0847
96 3916.91 0.0003 0.00002 43510 0.09002 11.1083 123.1529 11.0866
98 4653.68 0.0002 0.00002 51696 0.09002 11.1087 123.1963 11.0900
100 5529.04 0.0002 0.00002 61423 0.09002 11.1091 123.2335 11.0930

Compound Interest Factor Tables 595
10% TABLE 15 Discrete Cash Flow: Compound Interest Factors 10%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.1000 0.9091 1.00000 1.0000 1.10000 0.9091
2 1.2100 0.8264 0.47619 2.1000 0.57619 1.7355 0.8264 0.4762
3 1.3310 0.7513 0.30211 3.3100 0.40211 2.4869 2.3291 0.9366
4 1.4641 0.6830 0.21547 4.6410 0.31547 3.1699 4.3781 1.3812
5 1.6105 0.6209 0.16380 6.1051 0.26380 3.7908 6.8618 1.8101
6 1.7716 0.5645 0.12961 7.7156 0.22961 4.3553 9.6842 2.2236
7 1.9487 0.5132 0.10541 9.4872 0.20541 4.8684 12.7631 2.6216
8 2.1436 0.4665 0.08744 11.4359 0.18744 5.3349 16.0287 3.0045
9 2.3579 0.4241 0.07364 13.5795 0.17364 5.7590 19.4215 3.3724
10 2.5937 0.3855 0.06275 15.9374 0.16275 6.1446 22.8913 3.7255
11 2.8531 0.3505 0.05396 18.5312 0.15396 6.4951 26.3963 4.0641
12 3.1384 0.3186 0.04676 21.3843 0.14676 6.8137 29.9012 4.3884
13 3.4523 0.2897 0.04078 24.5227 0.14078 7.1034 33.3772 4.6988
14 3.7975 0.2633 0.03575 27.9750 0.13575 7.3667 36.8005 4.9955
15 4.1772 0.2394 0.03147 31.7725 0.13147 7.6061 40.1520 5.2789
16 4.5950 0.2176 0.02782 35.9497 0.12782 7.8237 43.4164 5.5493
17 5.0545 0.1978 0.02466 40.5447 0.12466 8.0216 46.5819 5.8071
18 5.5599 0.1799 0.02193 45.5992 0.12193 8.2014 49.6395 6.0526
19 6.1159 0.1635 0.01955 51.1591 0.11955 8.3649 52.5827 6.2861
20 6.7275 0.1486 0.01746 57.2750 0.11746 8.5136 55.4069 6.5081
21 7.4002 0.1351 0.01562 64.0025 0.11562 8.6487 58.1095 6.7189
22 8.1403 0.1228 0.01401 71.4027 0.11401 8.7715 60.6893 6.9189
23 8.9543 0.1117 0.01257 79.5430 0.11257 8.8832 63.1462 7.1085
24 9.8497 0.1015 0.01130 88.4973 0.11130 8.9847 65.4813 7.2881
25 10.8347 0.0923 0.01017 98.3471 0.11017 9.0770 67.6964 7.4580
26 11.9182 0.0839 0.00916 109.1818 0.10916 9.1609 69.7940 7.6186
27 13.1100 0.0763 0.00826 121.0999 0.10826 9.2372 71.7773 7.7704
28 14.4210 0.0693 0.00745 134.2099 0.10745 9.3066 73.6495 7.9137
29 15.8631 0.0630 0.00673 148.6309 0.10673 9.3696 75.4146 8.0489
30 17.4494 0.0573 0.00608 164.4940 0.10608 9.4269 77.0766 8.1762
31 19.1943 0.0521 0.00550 181.9434 0.10550 9.4790 78.6395 8.2962
32 21.1138 0.0474 0.00497 201.1378 0.10497 9.5264 80.1078 8.4091
33 23.2252 0.0431 0.00450 222.2515 0.10450 9.5694 81.4856 8.5152
34 25.5477 0.0391 0.00407 245.4767 0.10407 9.6086 82.7773 8.6149
35 28.1024 0.0356 0.00369 271.0244 0.10369 9.6442 83.9872 8.7086
40 45.2593 0.0221 0.00226 442.5926 0.10226 9.7791 88.9525 9.0962
45 72.8905 0.0137 0.00139 718.9048 0.10139 9.8628 92.4544 9.3740
50 117.3909 0.0085 0.00086 1163.91 0.10086 9.9148 94.8889 9.5704
55 189.0591 0.0053 0.00053 1880.59 0.10053 9.9471 96.5619 9.7075
60 304.4816 0.0033 0.00033 3034.82 0.10033 9.9672 97.7010 9.8023
65 490.3707 0.0020 0.00020 4893.71 0.10020 9.9796 98.4705 9.8672
70 789.7470 0.0013 0.00013 7887.47 0.10013 9.9873 98.9870 9.9113
75 1271.90 0.0008 0.00008 12709 0.10008 9.9921 99.3317 9.9410
80 2048.40 0.0005 0.00005 20474 0.10005 9.9951 99.5606 9.9609
85 3298.97 0.0003 0.00003 32980 0.10003 9.9970 99.7120 9.9742
90 5313.02 0.0002 0.00002 53120 0.10002 9.9981 99.8118 9.9831
95 8556.68 0.0001 0.00001 85557 0.10001 9.9988 99.8773 9.9889
96 9412.34 0.0001 0.00001 94113 0.10001 9.9989 99.8874 9.9898
98 11389 0.0001 0.00001 0.10001 9.9991 99.9052 9.9914
100 13781 0.0001 0.00001 0.10001 9.9993 99.9202 9.9927

596 Compound Interest Factor Tables
11% TABLE 16 Discrete Cash Flow: Compound Interest Factors 11%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.1100 0.9009 1.00000 1.0000 1.11000 0.9009
2 1.2321 0.8116 0.47393 2.1100 0.58393 1.7125 0.8116 0.4739
3 1.3676 0.7312 0.29921 3.3421 0.40921 2.4437 2.2740 0.9306
4 1.5181 0.6587 0.21233 4.7097 0.32233 3.1024 4.2502 1.3700
5 1.6851 0.5935 0.16057 6.2278 0.27057 3.6959 6.6240 1.7923
6 1.8704 0.5346 0.12638 7.9129 0.23638 4.2305 9.2972 2.1976
7 2.0762 0.4817 0.10222 9.7833 0.21222 4.7122 12.1872 2.5863
8 2.3045 0.4339 0.08432 11.8594 0.19432 5.1461 15.2246 2.9585
9 2.5580 0.3909 0.07060 14.1640 0.18060 5.5370 18.3520 3.3144
10 2.8394 0.3522 0.05980 16.7220 0.16980 5.8892 21.5217 3.6544
11 3.1518 0.3173 0.05112 19.5614 0.16112 6.2065 24.6945 3.9788
12 3.4985 0.2858 0.04403 22.7132 0.15403 6.4924 27.8388 4.2879
13 3.8833 0.2575 0.03815 26.2116 0.14815 6.7499 30.9290 4.5822
14 4.3104 0.2320 0.03323 30.0949 0.14323 6.9819 33.9449 4.8619
15 4.7846 0.2090 0.02907 34.4054 0.13907 7.1909 36.8709 5.1275
16 5.3109 0.1883 0.02552 39.1899 0.13552 7.3792 39.6953 5.3794
17 5.8951 0.1696 0.02247 44.5008 0.13247 7.5488 42.4095 5.6180
18 6.5436 0.1528 0.01984 50.3959 0.12984 7.7016 45.0074 5.8439
19 7.2633 0.1377 0.01756 56.9395 0.12756 7.8393 47.4856 6.0574
20 8.0623 0.1240 0.01558 64.2028 0.12558 7.9633 49.8423 6.2590
21 8.9492 0.1117 0.01384 72.2651 0.12384 8.0751 52.0771 6.4491
22 9.9336 0.1007 0.01231 81.2143 0.12231 8.1757 54.1912 6.6283
23 11.0263 0.0907 0.01097 91.1479 0.12097 8.2664 56.1864 6.7969
24 12.2392 0.0817 0.00979 102.1742 0.11979 8.3481 58.0656 6.9555
25 13.5855 0.0736 0.00874 114.4133 0.11874 8.4217 59.8322 7.1045
26 15.0799 0.0663 0.00781 127.9988 0.11781 8.4881 61.4900 7.2443
27 16.7386 0.0597 0.00699 143.0786 0.11699 8.5478 63.0433 7.3754
28 18.5799 0.0538 0.00626 159.8173 0.11626 8.6016 64.4965 7.4982
29 20.6237 0.0485 0.00561 178.3972 0.11561 8.6501 65.8542 7.6131
30 22.8923 0.0437 0.00502 199.0209 0.11502 8.6938 67.1210 7.7206
31 25.4104 0.0394 0.00451 221.9132 0.11451 8.7331 68.3016 7.8210
32 28.2056 0.0355 0.00404 247.3236 0.11404 8.7686 69.4007 7.9147
33 31.3082 0.0319 0.00363 275.5292 0.11363 8.8005 70.4228 8.0021
34 34.7521 0.0288 0.00326 306.8374 0.11326 8.8293 71.3724 8.0836
35 38.5749 0.0259 0.00293 341.5896 0.11293 8.8552 72.2538 8.1594
40 65.0009 0.0154 0.00172 581.8261 0.11172 8.9511 75.7789 8.4659
45 109.5302 0.0091 0.00101 986.6386 0.11101 9.0079 78.1551 8.6763
50 184.5648 0.0054 0.00060 1668.77 0.11060 9.0417 79.7341 8.8185
55 311.0025 0.0032 0.00035 2818.20 0.11035 9.0617 80.7712 8.9135
60 524.0572 0.0019 0.00021 4755.07 0.11021 9.0736 81.4461 8.9762
65 883.0669 0.0011 0.00012 8018.79 0.11012 9.0806 81.8819 9.0172
70 1488.02 0.0007 0.00007 13518 0.11007 9.0848 82.1614 9.0438
75 2507.40 0.0004 0.00004 22785 0.11004 9.0873 82.3397 9.0610
80 4225.11 0.0002 0.00003 38401 0.11003 9.0888 82.4529 9.0720
85 7119.56 0.0001 0.00002 64714 0.11002 9.0896 82.5245 9.0790

Compound Interest Factor Tables 597
12% TABLE 17 Discrete Cash Flow: Compound Interest Factors 12%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.1200 0.8929 1.00000 1.0000 1.12000 0.8929
2 1.2544 0.7972 0.47170 2.1200 0.59170 1.6901 0.7972 0.4717
3 1.4049 0.7118 0.29635 3.3744 0.41635 2.4018 2.2208 0.9246
4 1.5735 0.6355 0.20923 4.7793 0.32923 3.0373 4.1273 1.3589
5 1.7623 0.5674 0.15741 6.3528 0.27741 3.6048 6.3970 1.7746
6 1.9738 0.5066 0.12323 8.1152 0.24323 4.1114 8.9302 2.1720
7 2.2107 0.4523 0.09912 10.0890 0.21912 4.5638 11.6443 2.5512
8 2.4760 0.4039 0.08130 12.2997 0.20130 4.9676 14.4714 2.9131
9 2.7731 0.3606 0.06768 14.7757 0.18768 5.3282 17.3563 3.2574
10 3.1058 0.3220 0.05698 17.5487 0.17698 5.6502 20.2541 3.5847
11 3.4785 0.2875 0.04842 20.6546 0.16842 5.9377 23.1288 3.8953
12 3.8960 0.2567 0.04144 24.1331 0.16144 6.1944 25.9523 4.1897
13 4.3635 0.2292 0.03568 28.0291 0.15568 6.4235 28.7024 4.4683
14 4.8871 0.2046 0.03087 32.3926 0.15087 6.6282 31.3624 4.7317
15 5.4736 0.1827 0.02682 37.2797 0.14682 6.8109 33.9202 4.9803
16 6.1304 0.1631 0.02339 42.7533 0.14339 6.9740 36.3670 5.2147
17 6.8660 0.1456 0.02046 48.8837 0.14046 7.1196 38.6973 5.4353
18 7.6900 0.1300 0.01794 55.7497 0.13794 7.2497 40.9080 5.6427
19 8.6128 0.1161 0.01576 63.4397 0.13576 7.3658 42.9979 5.8375
20 9.6463 0.1037 0.01388 72.0524 0.13388 7.4694 44.9676 6.0202
21 10.8038 0.0926 0.01224 81.6987 0.13224 7.5620 46.8188 6.1913
22 12.1003 0.0826 0.01081 92.5026 0.13081 7.6446 48.5543 6.3514
23 13.5523 0.0738 0.00956 104.6029 0.12956 7.7184 50.1776 6.5010
24 15.1786 0.0659 0.00846 118.1552 0.12846 7.7843 51.6929 6.6406
25 17.0001 0.0588 0.00750 133.3339 0.12750 7.8431 53.1046 6.7708
26 19.0401 0.0525 0.00665 150.3339 0.12665 7.8957 54.4177 6.8921
27 21.3249 0.0469 0.00590 169.3740 0.12590 7.9426 55.6369 7.0049
28 23.8839 0.0419 0.00524 190.6989 0.12524 7.9844 56.7674 7.1098
29 26.7499 0.0374 0.00466 214.5828 0.12466 8.0218 57.8141 7.2071
30 29.9599 0.0334 0.00414 241.3327 0.12414 8.0552 58.7821 7.2974
31 33.5551 0.0298 0.00369 271.2926 0.12369 8.0850 59.6761 7.3811
32 37.5817 0.0266 0.00328 304.8477 0.12328 8.1116 60.5010 7.4586
33 42.0915 0.0238 0.00292 342.4294 0.12292 8.1354 61.2612 7.5302
34 47.1425 0.0212 0.00260 384.5210 0.12260 8.1566 61.9612 7.5965
35 52.7996 0.0189 0.00232 431.6635 0.12232 8.1755 62.6052 7.6577
40 93.0510 0.0107 0.00130 767.0914 0.12130 8.2438 65.1159 7.8988
45 163.9876 0.0061 0.0074 1358.23 0.12074 8.2825 66.7342 8.0572
50 289.0022 0.0035 0.00042 2400.02 0.12042 8.3045 67.7624 8.1597
55 509.3206 0.0020 0.00024 4236.01 0.12024 8.3170 68.4082 8.2251
60 897.5969 0.0011 0.00013 7471.64 0.12013 8.3240 68.8100 8.2664
65 1581.87 0.0006 0.00008 13174 0.12008 8.3281 69.0581 8.2922
70 2787.80 0.0004 0.00004 23223 0.12004 8.3303 69.2103 8.3082
75 4913.06 0.0002 0.00002 40934 0.12002 8.3316 69.3031 8.3181
80 8658.48 0.0001 0.00001 72146 0.12001 8.3324 69.3594 8.3241
85 15259 0.0001 0.00001 0.12001 8.3328 69.3935 8.3278

598 Compound Interest Factor Tables
14% TABLE 18 Discrete Cash Flow: Compound Interest Factors 14%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.1400 0.8772 1.00000 1.0000 1.14000 0.8772
2 1.2996 0.7695 0.46729 2.1400 0.60729 1.6467 0.7695 0.4673
3 1.4815 0.6750 0.29073 3.4396 0.43073 2.3216 2.1194 0.9129
4 1.6890 0.5921 0.20320 4.9211 0.34320 2.9137 3.8957 1.3370
5 1.9254 0.5194 0.15128 6.6101 0.29128 3.4331 5.9731 1.7399
6 2.1950 0.4556 0.11716 8.5355 0.25716 3.8887 8.2511 2.1218
7 2.5023 0.3996 0.09319 10.7305 0.23319 4.2883 10.6489 2.4832
8 2.8526 0.3506 0.07557 13.2328 0.21557 4.6389 13.1028 2.8246
9 3.2519 0.3075 0.06217 16.0853 0.20217 4.9464 15.5629 3.1463
10 3.7072 0.2697 0.05171 19.3373 0.19171 5.2161 17.9906 3.4490
11 4.2262 0.2366 0.04339 23.0445 0.18339 5.4527 20.3567 3.7333
12 4.8179 0.2076 0.03667 27.2707 0.17667 5.6603 22.6399 3.9998
13 5.4924 0.1821 0.03116 32.0887 0.17116 5.8424 24.8247 4.2491
14 6.2613 0.1597 0.02661 37.5811 0.16661 6.0021 26.9009 4.4819
15 7.1379 0.1401 0.02281 43.8424 0.16281 6.1422 28.8623 4.6990
16 8.1372 0.1229 0.01962 50.9804 0.15962 6.2651 30.7057 4.9011
17 9.2765 0.1078 0.01692 59.1176 0.15692 6.3729 32.4305 5.0888
18 10.5752 0.0946 0.01462 68.3941 0.15462 6.4674 34.0380 5.2630
19 12.0557 0.0829 0.01266 78.9692 0.15266 6.5504 35.5311 5.4243
20 13.7435 0.0728 0.01099 91.0249 0.15099 6.6231 36.9135 5.5734
21 15.6676 0.0638 0.00954 104.7684 0.14954 6.6870 38.1901 5.7111
22 17.8610 0.0560 0.00830 120.4360 0.14830 6.7429 39.3658 5.8381
23 20.3616 0.0491 0.00723 138.2970 0.14723 6.7921 40.4463 5.9549
24 23.2122 0.0431 0.00630 158.6586 0.14630 6.8351 41.4371 6.0624
25 26.4619 0.0378 0.00550 181.8708 0.14550 6.8729 42.3441 6.1610
26 30.1666 0.0331 0.00480 208.3327 0.14480 6.9061 43.1728 6.2514
27 34.3899 0.0291 0.00419 238.4993 0.14419 6.9352 43.9289 6.3342
28 39.2045 0.0255 0.00366 272.8892 0.14366 6.9607 44.6176 6.4100
29 44.6931 0.0224 0.00320 312.0937 0.14320 6.9830 45.2441 6.4791
30 50.9502 0.0196 0.00280 356.7868 0.14280 7.0027 45.8132 6.5423
31 58.0832 0.0172 0.00245 407.7370 0.14245 7.0199 46.3297 6.5998
32 66.2148 0.0151 0.00215 465.8202 0.14215 7.0350 46.7979 6.6522
33 75.4849 0.0132 0.00188 532.0350 0.14188 7.0482 47.2218 6.6998
34 86.0528 0.0116 0.00165 607.5199 0.14165 7.0599 47.6053 6.7431
35 98.1002 0.0102 0.00144 693.5727 0.14144 7.0700 47.9519 6.7824
40 188.8835 0.0053 0.00075 1342.03 0.14075 7.1050 49.2376 6.9300
45 363.6791 0.0027 0.00039 2590.56 0.14039 7.1232 49.9963 7.0188
50 700.2330 0.0014 0.00020 4994.52 0.14020 7.1327 50.4375 7.0714
55 1348.24 0.0007 0.00010 9623.13 0.14010 7.1376 50.6912 7.1020
60 2595.92 0.0004 0.00005 18535 0.14005 7.1401 50.8357 7.1197
65 4998.22 0.0002 0.00003 35694 0.14003 7.1414 50.9173 7.1298
70 9623.64 0.0001 0.00001 68733 0.14001 7.1421 50.9632 7.1356
75 18530 0.0001 0.00001 0.14001 7.1425 50.9887 7.1388
80 35677 0.14000 7.1427 51.0030 7.1406
85 68693 0.14000 7.1428 51.0108 7.1416

Compound Interest Factor Tables 599
15% TABLE 19 Discrete Cash Flow: Compound Interest Factors 15%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.1500 0.8696 1.00000 1.0000 1.15000 0.8696
2 1.3225 0.7561 0.46512 2.1500 0.61512 1.6257 0.7561 0.4651
3 1.5209 0.6575 0.28798 3.4725 0.43798 2.2832 2.0712 0.9071
4 1.7490 0.5718 0.20027 4.9934 0.35027 2.8550 3.7864 1.3263
5 2.0114 0.4972 0.14832 6.7424 0.29832 3.3522 5.7751 1.7228
6 2.3131 0.4323 0.11424 8.7537 0.26424 3.7845 7.9368 2.0972
7 2.6600 0.3759 0.09036 11.0668 0.24036 4.1604 10.1924 2.4498
8 3.0590 0.3269 0.07285 13.7268 0.22285 4.4873 12.4807 2.7813
9 3.5179 0.2843 0.05957 16.7858 0.20957 4.7716 14.7548 3.0922
10 4.0456 0.2472 0.04925 20.3037 0.19925 5.0188 16.9795 3.3832
11 4.6524 0.2149 0.04107 24.3493 0.19107 5.2337 19.1289 3.6549
12 5.3503 0.1869 0.03448 29.0017 0.18448 5.4206 21.1849 3.9082
13 6.1528 0.1625 0.02911 34.3519 0.17911 5.5831 23.1352 4.1438
14 7.0757 0.1413 0.02469 40.5047 0.17469 5.7245 24.9725 4.3624
15 8.1371 0.1229 0.02102 47.5804 0.17102 5.8474 26.6930 4.5650
16 9.3576 0.1069 0.01795 55.7175 0.16795 5.9542 28.2960 4.7522
17 10.7613 0.0929 0.01537 65.0751 0.16537 6.0472 29.7828 4.9251
18 12.3755 0.0808 0.01319 75.8364 0.16319 6.1280 31.1565 5.0843
19 14.2318 0.0703 0.01134 88.2118 0.16134 6.1982 32.4213 5.2307
20 16.3665 0.0611 0.00976 102.4436 0.15976 6.2593 33.5822 5.3651
21 18.8215 0.0531 0.00842 118.8101 0.15842 6.3125 34.6448 5.4883
22 21.6447 0.0462 0.00727 137.6316 0.15727 6.3587 35.6150 5.6010
23 24.8915 0.0402 0.00628 159.2764 0.15628 6.3988 36.4988 5.7040
24 28.6252 0.0349 0.00543 184.1678 0.15543 6.4338 37.3023 5.7979
25 32.9190 0.0304 0.00470 212.7930 0.15470 6.4641 38.0314 5.8834
26 37.8568 0.0264 0.00407 245.7120 0.15407 6.4906 38.6918 5.9612
27 43.5353 0.0230 0.00353 283.5688 0.15353 6.5135 39.2890 6.0319
28 50.0656 0.0200 0.00306 327.1041 0.15306 6.5335 39.8283 6.0960
29 57.5755 0.0174 0.00265 377.1697 0.15265 6.5509 40.3146 6.1541
30 66.2118 0.0151 0.00230 434.7451 0.15230 6.5660 40.7526 6.2066
31 76.1435 0.0131 0.00200 500.9569 0.15200 6.5791 41.1466 6.2541
32 87.5651 0.0114 0.00173 577.1005 0.15173 6.5905 41.5006 6.2970
33 100.6998 0.0099 0.00150 664.6655 0.15150 6.6005 41.8184 6.3357
34 115.8048 0.0086 0.00131 765.3654 0.15131 6.6091 42.1033 6.3705
35 133.1755 0.0075 0.00113 881.1702 0.15113 6.6166 42.3586 6.4019
40 267.8635 0.0037 0.00056 1779.09 0.15056 6.6418 43.2830 6.5168
45 538.7693 0.0019 0.00028 3585.13 0.15028 6.6543 43.8051 6.5830
50 1083.66 0.0009 0.00014 7217.72 0.15014 6.6605 44.0958 6.6205
55 2179.62 0.0005 0.00007 14524 0.15007 6.6636 44.2558 6.6414
60 4384.00 0.0002 0.00003 29220 0.15003 6.6651 44.3431 6.6530
65 8817.79 0.0001 0.00002 58779 0.15002 6.6659 44.3903 6.6593
70 17736 0.0001 0.00001 0.15001 6.6663 44.4156 6.6627
75 35673 0.15000 6.6665 44.4292 6.6646
80 71751 0.15000 6.6666 44.4364 6.6656
85 0.15000 6.6666 44.4402 6.6661

600 Compound Interest Factor Tables
16% TABLE 20 Discrete Cash Flow: Compound Interest Factors 16%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.1600 0.8621 1.00000 1.0000 1.16000 0.8621
2 1.3456 0.7432 0.46296 2.1600 0.62296 1.6052 0.7432 0.4630
3 1.5609 0.6407 0.28526 3.5056 0.44526 2.2459 2.0245 0.9014
4 1.8106 0.5523 0.19738 5.0665 0.35738 2.7982 3.6814 1.3156
5 2.1003 0.4761 0.14541 6.8771 0.30541 3.2743 5.5858 1.7060
6 2.4364 0.4104 0.11139 8.9775 0.27139 3.6847 7.6380 2.0729
7 2.8262 0.3538 0.08761 11.4139 0.24761 4.0386 9.7610 2.4169
8 3.2784 0.3050 0.07022 14.2401 0.23022 4.3436 11.8962 2.7388
9 3.8030 0.2630 0.05708 17.5185 0.21708 4.6065 13.9998 3.0391
10 4.4114 0.2267 0.04690 21.3215 0.20690 4.8332 16.0399 3.3187
11 5.1173 0.1954 0.03886 25.7329 0.19886 5.0286 17.9941 3.5783
12 5.9360 0.1685 0.03241 30.8502 0.19241 5.1971 19.8472 3.8189
13 6.8858 0.1452 0.02718 36.7862 0.18718 5.3423 21.5899 4.0413
14 7.9875 0.1252 0.02290 43.6720 0.18290 5.4675 23.2175 4.2464
15 9.2655 0.1079 0.01936 51.6595 0.17936 5.5755 24.7284 4.4352
16 10.7480 0.0930 0.01641 60.9250 0.17641 5.6685 26.1241 4.6086
17 12.4677 0.0802 0.01395 71.6730 0.17395 5.7487 27.4074 4.7676
18 14.4625 0.0691 0.01188 84.1407 0.17188 5.8178 28.5828 4.9130
19 16.7765 0.0596 0.01014 98.6032 0.17014 5.8775 29.6557 5.0457
20 19.4608 0.0514 0.00867 115.3797 0.16867 5.9288 30.6321 5.1666
22 26.1864 0.0382 0.00635 157.4150 0.16635 6.0113 32.3200 5.3765
24 35.2364 0.0284 0.00467 213.9776 0.16467 6.0726 33.6970 5.5490
26 47.4141 0.0211 0.00345 290.0883 0.16345 6.1182 34.8114 5.6898
28 63.8004 0.0157 0.00255 392.5028 0.16255 6.1520 35.7073 5.8041
30 85.8499 0.0116 0.00189 530.3117 0.16189 6.1772 36.4234 5.8964
32 115.5196 0.0087 0.00140 715.7475 0.16140 6.1959 36.9930 5.9706
34 155.4432 0.0064 0.00104 965.2698 0.16104 6.2098 37.4441 6.0299
35 180.3141 0.0055 0.00089 1120.71 0.16089 6.2153 37.6327 6.0548
36 209.1643 0.0048 0.00077 1301.03 0.16077 6.2201 37.8000 6.0771
38 281.4515 0.0036 0.00057 1752.82 0.16057 6.2278 38.0799 6.1145
40 378.7212 0.0026 0.00042 2360.76 0.16042 6.2335 38.2992 6.1441
45 795.4438 0.0013 0.00020 4965.27 0.16020 6.2421 38.6598 6.1934
50 1670.70 0.0006 0.00010 10436 0.16010 6.2463 38.8521 6.2201
55 3509.05 0.0003 0.00005 21925 0.16005 6.2482 38.9534 6.2343
60 7370.20 0.0001 0.00002 46058 0.16002 6.2492 39.0063 6.2419

Compound Interest Factor Tables 601
18% TABLE 21 Discrete Cash Flow: Compound Interest Factors 18%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.1800 0.8475 1.00000 1.0000 1.18000 0.8475
2 1.3924 0.7182 0.45872 2.1800 0.63872 1.5656 0.7182 0.4587
3 1.6430 0.6086 0.27992 3.5724 0.45992 2.1743 1.9354 0.8902
4 1.9388 0.5158 0.19174 5.2154 0.37174 2.6901 3.4828 1.2947
5 2.2878 0.4371 0.13978 7.1542 0.31978 3.1272 5.2312 1.6728
6 2.6996 0.3704 0.10591 9.4420 0.28591 3.4976 7.0834 2.0252
7 3.1855 0.3139 0.08236 12.1415 0.26236 3.8115 8.9670 2.3526
8 3.7589 0.2660 0.06524 15.3270 0.24524 4.0776 10.8292 2.6558
9 4.4355 0.2255 0.05239 19.0859 0.23239 4.3030 12.6329 2.9358
10 5.2338 0.1911 0.04251 23.5213 0.22251 4.4941 14.3525 3.1936
11 6.1759 0.1619 0.03478 28.7551 0.21478 4.6560 15.9716 3.4303
12 7.2876 0.1372 0.02863 34.9311 0.20863 4.7932 17.4811 3.6470
13 8.5994 0.1163 0.02369 42.2187 0.20369 4.9095 18.8765 3.8449
14 10.1472 0.0985 0.01968 50.8180 0.19968 5.0081 20.1576 4.0250
15 11.9737 0.0835 0.01640 60.9653 0.19640 5.0916 21.3269 4.1887
16 14.1290 0.0708 0.01371 72.9390 0.19371 5.1624 22.3885 4.3369
17 16.6722 0.0600 0.01149 87.0680 0.19149 5.2223 23.3482 4.4708
18 19.6733 0.0508 0.00964 103.7403 0.18964 5.2732 24.2123 4.5916
19 23.2144 0.0431 0.00810 123.4135 0.18810 5.3162 24.9877 4.7003
20 27.3930 0.0365 0.00682 146.6280 0.18682 5.3527 25.6813 4.7978
22 38.1421 0.0262 0.00485 206.3448 0.18485 5.4099 26.8506 4.9632
24 53.1090 0.0188 0.00345 289.4945 0.18345 5.4509 27.7725 5.0950
26 73.9490 0.0135 0.00247 405.2721 0.18247 5.4804 28.4935 5.1991
28 102.9666 0.0097 0.00177 566.4809 0.18177 5.5016 29.0537 5.2810
30 143.3706 0.0070 0.00126 790.9480 0.18126 5.5168 29.4864 5.3448
32 199.6293 0.0050 0.00091 1103.50 0.18091 5.5277 29.8191 5.3945
34 277.9638 0.0036 0.00065 1538.69 0.18065 5.5356 30.0736 5.4328
35 327.9973 0.0030 0.00055 1816.65 0.18055 5.5386 30.1773 5.4485
36 387.0368 0.0026 0.00047 2144.65 0.18047 5.5412 30.2677 5.4623
38 538.9100 0.0019 0.00033 2988.39 0.18033 5.5452 30.4152 5.4849
40 750.3783 0.0013 0.00024 4163.21 0.18024 5.5482 30.5269 5.5022
45 1716.68 0.0006 0.00010 9531.58 0.18010 5.5523 30.7006 5.5293
50 3927.36 0.0003 0.00005 21813 0.18005 5.5541 30.7856 5.5428
55 8984.84 0.0001 0.00002 49910 0.18002 5.5549 30.8268 5.5494
60 20555 114190 0.18001 5.5553 30.8465 5.5526

602 Compound Interest Factor Tables
20% TABLE 22 Discrete Cash Flow: Compound Interest Factors 20%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.2000 0.8333 1.00000 1.0000 1.20000 0.8333
2 1.4400 0.6944 0.45455 2.2000 0.65455 1.5278 0.6944 0.4545
3 1.7280 0.5787 0.27473 3.6400 0.47473 2.1065 1.8519 0.8791
4 2.0736 0.4823 0.18629 5.3680 0.38629 2.5887 3.2986 1.2742
5 2.4883 0.4019 0.13438 7.4416 0.33438 2.9906 4.9061 1.6405
6 2.9860 0.3349 0.10071 9.9299 0.30071 3.3255 6.5806 1.9788
7 3.5832 0.2791 0.07742 12.9159 0.27742 3.6046 8.2551 2.2902
8 4.2998 0.2326 0.06061 16.4991 0.26061 3.8372 9.8831 2.5756
9 5.1598 0.1938 0.04808 20.7989 0.24808 4.0310 11.4335 2.8364
10 6.1917 0.1615 0.03852 25.9587 0.23852 4.1925 12.8871 3.0739
11 7.4301 0.1346 0.03110 32.1504 0.23110 4.3271 14.2330 3.2893
12 8.9161 0.1122 0.02526 39.5805 0.22526 4.4392 15.4667 3.4841
13 10.6993 0.0935 0.02062 48.4966 0.22062 4.5327 16.5883 3.6597
14 12.8392 0.0779 0.01689 59.1959 0.21689 4.6106 17.6008 3.8175
15 15.4070 0.0649 0.01388 72.0351 0.21388 4.6755 18.5095 3.9588
16 18.4884 0.0541 0.01144 87.4421 0.21144 4.7296 19.3208 4.0851
17 22.1861 0.0451 0.00944 105.9306 0.20944 4.7746 20.0419 4.1976
18 26.6233 0.0376 0.00781 128.1167 0.20781 4.8122 20.6805 4.2975
19 31.9480 0.0313 0.00646 154.7400 0.20646 4.8435 21.2439 4.3861
20 38.3376 0.0261 0.00536 186.6880 0.20536 4.8696 21.7395 4.4643
22 55.2061 0.0181 0.00369 271.0307 0.20369 4.9094 22.5546 4.5941
24 79.4968 0.0126 0.00255 392.4842 0.20255 4.9371 23.1760 4.6943
26 114.4755 0.0087 0.00176 567.3773 0.20176 4.9563 23.6460 4.7709
28 164.8447 0.0061 0.00122 819.2233 0.20122 4.9697 23.9991 4.8291
30 237.3763 0.0042 0.00085 1181.88 0.20085 4.9789 24.2628 4.8731
32 341.8219 0.0029 0.00059 1704.11 0.20059 4.9854 24.4588 4.9061
34 492.2235 0.0020 0.00041 2456.12 0.20041 4.9898 24.6038 4.9308
35 590.6682 0.0017 0.00034 2948.34 0.20034 4.9915 24.6614 4.9406
36 708.8019 0.0014 0.00028 3539.01 0.20028 4.9929 24.7108 4.9491
38 1020.67 0.0010 0.00020 5098.37 0.20020 4.9951 24.7894 4.9627
40 1469.77 0.0007 0.00014 7343.86 0.20014 4.9966 24.8469 4.9728
45 3657.26 0.0003 0.00005 18281 0.20005 4.9986 24.9316 4.9877
50 9100.44 0.0001 0.00002 45497 0.20002 4.9995 24.9698 4.9945
55 22645 0.00001 0.20001 4.9998 24.9868 4.9976

Compound Interest Factor Tables 603
22% TABLE 23 Discrete Cash Flow: Compound Interest Factors 22%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.2200 0.8197 1.00000 1.0000 1.22000 0.8197
2 1.4884 0.6719 0.45045 2.2200 0.67045 1.4915 0.6719 0.4505
3 1.8158 0.5507 0.26966 3.7084 0.48966 2.0422 1.7733 0.8683
4 2.2153 0.4514 0.18102 5.5242 0.40102 2.4936 3.1275 1.2542
5 2.7027 0.3700 0.12921 7.7396 0.34921 2.8636 4.6075 1.6090
6 3.2973 0.3033 0.09576 10.4423 0.31576 3.1669 6.1239 1.9337
7 4.0227 0.2486 0.07278 13.7396 0.29278 3.4155 7.6154 2.2297
8 4.9077 0.2038 0.05630 17.7623 0.27630 3.6193 9.0417 2.4982
9 5.9874 0.1670 0.04411 22.6700 0.26411 3.7863 10.3779 2.7409
10 7.3046 0.1369 0.03489 28.6574 0.25489 3.9232 11.6100 2.9593
11 8.9117 0.1122 0.02781 35.9620 0.24781 4.0354 12.7321 3.1551
12 10.8722 0.0920 0.02228 44.8737 0.24228 4.1274 13.7438 3.3299
13 13.2641 0.0754 0.01794 55.7459 0.23794 4.2028 14.6485 3.4855
14 16.1822 0.0618 0.01449 69.0100 0.23449 4.2646 15.4519 3.6233
15 19.7423 0.0507 0.01174 85.1922 0.23174 4.3152 16.1610 3.7451
16 24.0856 0.0415 0.00953 104.9345 0.22953 4.3567 16.7838 3.8524
17 29.3844 0.0340 0.00775 129.0201 0.22775 4.3908 17.3283 3.9465
18 35.8490 0.0279 0.00631 158.4045 0.22631 4.4187 17.8025 4.0289
19 43.7358 0.0229 0.00515 194.2535 0.22515 4.4415 18.2141 4.1009
20 53.3576 0.0187 0.00420 237.9893 0.22420 4.4603 18.5702 4.1635
22 79.4175 0.0126 0.00281 356.4432 0.22281 4.4882 19.1418 4.2649
24 118.2050 0.0085 0.00188 532.7501 0.22188 4.5070 19.5635 4.3407
26 175.9364 0.0057 0.00126 795.1653 0.22126 4.5196 19.8720 4.3968
28 261.8637 0.0038 0.00084 1185.74 0.22084 4.5281 20.0962 4.4381
30 389.7579 0.0026 0.00057 1767.08 0.22057 4.5338 20.2583 4.4683
32 580.1156 0.0017 0.00038 2632.34 0.22038 4.5376 20.3748 4.4902
34 863.4441 0.0012 0.00026 3920.20 0.22026 4.5402 20.4582 4.5060
35 1053.40 0.0009 0.00021 4783.64 0.22021 4.5411 20.4905 4.5122
36 1285.15 0.0008 0.00017 5837.05 0.22017 4.5419 20.5178 4.5174
38 1912.82 0.0005 0.00012 8690.08 0.22012 4.5431 20.5601 4.5256
40 2847.04 0.0004 0.00008 12937 0.22008 4.5439 20.5900 4.5314
45 7694.71 0.0001 0.00003 34971 0.22003 4.5449 20.6319 4.5396
50 20797 0.00001 94525 0.22001 4.5452 20.6492 4.5431
55 56207 0.22000 4.5454 20.6563 4.5445

604 Compound Interest Factor Tables
24% TABLE 24 Discrete Cash Flow: Compound Interest Factors 24%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.2400 0.8065 1.00000 1.0000 1.24000 0.8065
2 1.5376 0.6504 0.44643 2.2400 0.68643 1.4568 0.6504 0.4464
3 1.9066 0.5245 0.26472 3.7776 0.50472 1.9813 1.6993 0.8577
4 2.3642 0.4230 0.17593 5.6842 0.41593 2.4043 2.9683 1.2346
5 2.9316 0.3411 0.12425 8.0484 0.36425 2.7454 4.3327 1.5782
6 3.6352 0.2751 0.09107 10.9801 0.33107 3.0205 5.7081 1.8898
7 4.5077 0.2218 0.06842 14.6153 0.30842 3.2423 7.0392 2.1710
8 5.5895 0.1789 0.05229 19.1229 0.29229 3.4212 8.2915 2.4236
9 6.9310 0.1443 0.04047 24.7125 0.28047 3.5655 9.4458 2.6492
10 8.5944 0.1164 0.03160 31.6434 0.27160 3.6819 10.4930 2.8499
11 10.6571 0.0938 0.02485 40.2379 0.26485 3.7757 11.4313 3.0276
12 13.2148 0.0757 0.01965 50.8950 0.25965 3.8514 12.2637 3.1843
13 16.3863 0.0610 0.01560 64.1097 0.25560 3.9124 12.9960 3.3218
14 20.3191 0.0492 0.01242 80.4961 0.25242 3.9616 13.6358 3.4420
15 25.1956 0.0397 0.00992 100.8151 0.24992 4.0013 14.1915 3.5467
16 31.2426 0.0320 0.00794 126.0108 0.24794 4.0333 14.6716 3.6376
17 38.7408 0.0258 0.00636 157.2534 0.24636 4.0591 15.0846 3.7162
18 48.0386 0.0208 0.00510 195.9942 0.24510 4.0799 15.4385 3.7840
19 59.5679 0.0168 0.00410 244.0328 0.24410 4.0967 15.7406 3.8423
20 73.8641 0.0135 0.00329 303.6006 0.24329 4.1103 15.9979 3.8922
22 113.5735 0.0088 0.00213 469.0563 0.24213 4.1300 16.4011 3.9712
24 174.6306 0.0057 0.00138 723.4610 0.24138 4.1428 16.6891 4.0284
26 268.5121 0.0037 0.00090 1114.63 0.24090 4.1511 16.8930 4.0695
28 412.8642 0.0024 0.00058 1716.10 0.24058 4.1566 17.0365 4.0987
30 634.8199 0.0016 0.00038 2640.92 0.24038 4.1601 17.1369 4.1193
32 976.0991 0.0010 0.00025 4062.91 0.24025 4.1624 17.2067 4.1338
34 1500.85 0.0007 0.00016 6249.38 0.24016 4.1639 17.2552 4.1440
35 1861.05 0.0005 0.00013 7750.23 0.24013 4.1664 17.2734 4.1479
36 2307.71 0.0004 0.00010 9611.28 0.24010 4.1649 17.2886 4.1511
38 3548.33 0.0003 0.00007 14781 0.24007 4.1655 17.3116 4.1560
40 5455.91 0.0002 0.00004 22729 0.24004 4.1659 17.3274 4.1593
45 15995 0.0001 0.00002 66640 0.24002 4.1664 17.3483 4.1639
50 46890 0.00001 0.24001 4.1666 17.3563 4.1653
55 0.24000 4.1666 17.3593 4.1663

Compound Interest Factor Tables 605
25% TABLE 25 Discrete Cash Flow: Compound Interest Factors 25%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.2500 0.8000 1.00000 1.0000 1.25000 0.8000
2 1.5625 0.6400 0.44444 2.2500 0.69444 1.4400 0.6400 0.4444
3 1.9531 0.5120 0.26230 3.8125 0.51230 1.9520 1.6640 0.8525
4 2.4414 0.4096 0.17344 5.7656 0.42344 2.3616 2.8928 1.2249
5 3.0518 0.3277 0.12185 8.2070 0.37185 2.6893 4.2035 1.5631
6 3.8147 0.2621 0.08882 11.2588 0.33882 2.9514 5.5142 1.8683
7 4.7684 0.2097 0.06634 15.0735 0.31634 3.1611 6.7725 2.1424
8 5.9605 0.1678 0.05040 19.8419 0.30040 3.3289 7.9469 2.3872
9 7.4506 0.1342 0.03876 25.8023 0.28876 3.4631 9.0207 2.6048
10 9.3132 0.1074 0.03007 33.2529 0.28007 3.5705 9.9870 2.7971
11 11.6415 0.0859 0.02349 42.5661 0.27349 3.6564 10.8460 2.9663
12 14.5519 0.0687 0.01845 54.2077 0.26845 3.7251 11.6020 3.1145
13 18.1899 0.0550 0.01454 68.7596 0.26454 3.7801 12.2617 3.2437
14 22.7374 0.0440 0.01150 86.9495 0.26150 3.8241 12.8334 3.3559
15 28.4217 0.0352 0.00912 109.6868 0.25912 3.8593 13.3260 3.4530
16 35.5271 0.0281 0.00724 138.1085 0.25724 3.8874 13.7482 3.5366
17 44.4089 0.0225 0.00576 173.6357 0.25576 3.9099 14.1085 3.6084
18 55.5112 0.0180 0.00459 218.0446 0.25459 3.9279 14.4147 3.6698
19 69.3889 0.0144 0.00366 273.5558 0.25366 3.9424 14.6741 3.7222
20 86.7362 0.0115 0.00292 342.9447 0.25292 3.9539 14.8932 3.7667
22 135.5253 0.0074 0.00186 538.1011 0.25186 3.9705 15.2326 3.8365
24 211.7582 0.0047 0.00119 843.0329 0.25119 3.9811 15.4711 3.8861
26 330.8722 0.0030 0.00076 1319.49 0.25076 3.9879 15.6373 3.9212
28 516.9879 0.0019 0.00048 2063.95 0.25048 3.9923 15.7524 3.9457
30 807.7936 0.0012 0.00031 3227.17 0.25031 3.9950 15.8316 3.9628
32 1262.18 0.0008 0.00020 5044.71 0.25020 3.9968 15.8859 3.9746
34 1972.15 0.0005 0.00013 7884.61 0.25013 3.9980 15.9229 3.9828
35 2465.19 0.0004 0.00010 9856.76 .025010 3.9984 15.9367 3.9858
36 3081.49 0.0003 0.00008 12322 0.25008 3.9987 15.9481 3.9883
38 4814.82 0.0002 0.00005 19255 0.25005 3.9992 15.9651 3.9921
40 7523.16 0.0001 0.00003 30089 0.25003 3.9995 15.9766 3.9947
45 22959 0.00001 91831 0.25001 3.9998 15.9915 3.9980
50 70065 0.25000 3.9999 15.9969 3.9993
55 0.25000 4.0000 15.9989 3.9997

606 Compound Interest Factor Tables
30% TABLE 26 Discrete Cash Flow: Compound Interest Factors 30%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.3000 0.7692 1.00000 1.0000 1.30000 0.7692
2 1.6900 0.5917 0.43478 2.3000 0.73478 1.3609 0.5917 0.4348
3 2.1970 0.4552 0.25063 3.9900 0.55063 1.8161 1.5020 0.8271
4 2.8561 0.3501 0.16163 6.1870 0.46163 2.1662 2.5524 1.1783
5 3.7129 0.2693 0.11058 9.0431 0.41058 2.4356 3.6297 1.4903
6 4.8268 0.2072 0.07839 12.7560 0.37839 2.6427 4.6656 1.7654
7 6.2749 0.1594 0.05687 17.5828 0.35687 2.8021 5.6218 2.0063
8 8.1573 0.1226 0.04192 23.8577 0.34192 2.9247 6.4800 2.2156
9 10.6045 0.0943 0.03124 32.0150 0.33124 3.0190 7.2343 2.3963
10 13.7858 0.0725 0.02346 42.6195 0.32346 3.0915 7.8872 2.5512
11 17.9216 0.0558 0.01773 56.4053 0.31773 3.1473 8.4452 2.6833
12 23.2981 0.0429 0.01345 74.3270 0.31345 3.1903 8.9173 2.7952
13 30.2875 0.0330 0.01024 97.6250 0.31024 3.2233 9.3135 2.8895
14 39.3738 0.0254 0.00782 127.9125 0.30782 3.2487 9.6437 2.9685
15 51.1859 0.0195 0.00598 167.2863 0.30598 3.2682 9.9172 3.0344
16 66.5417 0.0150 0.00458 218.4722 0.30458 3.2832 10.1426 3.0892
17 86.5042 0.0116 0.00351 285.0139 0.30351 3.2948 10.3276 3.1345
18 112.4554 0.0089 0.00269 371.5180 0.30269 3.3037 10.4788 3.1718
19 146.1920 0.0068 0.00207 483.9734 0.30207 3.3105 10.6019 3.2025
20 190.0496 0.0053 0.00159 630.1655 0.30159 3.3158 10.7019 3.2275
22 321.1839 0.0031 0.00094 1067.28 0.30094 3.3230 10.8482 3.2646
24 542.8008 0.0018 0.00055 1806.00 0.30055 3.3272 10.9433 3.2890
25 705.6410 0.0014 0.00043 2348.80 0.30043 3.3286 10.9773 3.2979
26 917.3333 0.0011 0.00033 3054.44 0.30033 3.3297 11.0045 3.3050
28 1550.29 0.0006 0.00019 5164.31 0.30019 3.3312 11.0437 3.3153
30 2620.00 0.0004 0.00011 8729.99 0.30011 3.3321 11.0687 3.3219
32 4427.79 0.0002 0.00007 14756 0.30007 3.3326 11.0845 3.3261
34 7482.97 0.0001 0.00004 24940 0.30004 3.3329 11.0945 3.3288
35 9727.86 0.0001 0.00003 32423 0.30003 3.3330 11.0980 3.3297

Compound Interest Factor Tables 607
35% TABLE 27 Discrete Cash Flow: Compound Interest Factors 35%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.3500 0.7407 1.00000 1.0000 1.35000 0.7407
2 1.8225 0.5487 0.42553 2.3500 0.77553 1.2894 0.5487 0.4255
3 2.4604 0.4064 0.23966 4.1725 0.58966 1.6959 1.3616 0.8029
4 3.3215 0.3011 0.15076 6.6329 0.50076 1.9969 2.2648 1.1341
5 4.4840 0.2230 0.10046 9.9544 0.45046 2.2200 3.1568 1.4220
6 6.0534 0.1652 0.06926 14.4384 0.41926 2.3852 3.9828 1.6698
7 8.1722 0.1224 0.04880 20.4919 0.39880 2.5075 4.7170 1.8811
8 11.0324 0.0906 0.03489 28.6640 0.38489 2.5982 5.3515 2.0597
9 14.8937 0.0671 0.02519 39.6964 0.37519 2.6653 5.8886 2.2094
10 20.1066 0.0497 0.01832 54.5902 0.36832 2.7150 6.3363 2.3338
11 27.1439 0.0368 0.01339 74.6967 0.36339 2.7519 6.7047 2.4364
12 36.6442 0.0273 0.00982 101.8406 0.35982 2.7792 7.0049 2.5205
13 49.4697 0.0202 0.00722 138.4848 0.35722 2.7994 7.2474 2.5889
14 66.7841 0.0150 0.00532 187.9544 0.35532 2.8144 7.4421 2.6443
15 90.1585 0.0111 0.00393 254.7385 0.35393 2.8255 7.5974 2.6889
16 121.7139 0.0082 0.00290 344.8970 0.35290 2.8337 7.7206 2.7246
17 164.3138 0.0061 0.00214 466.6109 0.35214 2.8398 7.8180 2.7530
18 221.8236 0.0045 0.00158 630.9247 0.35158 2.8443 7.8946 2.7756
19 299.4619 0.0033 0.00117 852.7483 0.35117 2.8476 7.9547 2.7935
20 404.2736 0.0025 0.00087 1152.21 0.35087 2.8501 8.0017 2.8075
22 736.7886 0.0014 0.00048 2102.25 0.35048 2.8533 8.0669 2.8272
24 1342.80 0.0007 0.00026 3833.71 0.35026 2.8550 8.1061 2.8393
25 1812.78 0.0006 0.00019 5176.50 0.35019 2.8556 8.1194 2.8433
26 2447.25 0.0004 0.00014 6989.28 0.35014 2.8560 8.1296 2.8465
28 4460.11 0.0002 0.00008 12740 0.35008 2.8565 8.1435 2.8509
30 8128.55 0.0001 0.00004 23222 0.35004 2.8568 8.1517 2.8535
32 14814 0.0001 0.00002 42324 0.35002 2.8569 8.1565 2.8550
34 26999 0.00001 77137 0.35001 2.8570 8.1594 2.8559
35 36449 0.00001 0.35001 2.8571 8.1603 2.8562

608 Compound Interest Factor Tables
40% TABLE 28 Discrete Cash Flow: Compound Interest Factors 40%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.4000 0.7143 1.00000 1.0000 1.40000 0.7143
2 1.9600 0.5102 0.41667 2.4000 0.81667 1.2245 0.5102 0.4167
3 2.7440 0.3644 0.22936 4.3600 0.62936 1.5889 1.2391 0.7798
4 3.8416 0.2603 0.14077 7.1040 0.54077 1.8492 2.0200 1.0923
5 5.3782 0.1859 0.09136 10.9456 0.49136 2.0352 2.7637 1.3580
6 7.5295 0.1328 0.06126 16.3238 0.46126 2.1680 3.4278 1.5811
7 10.5414 0.0949 0.04192 23.8534 0.44192 2.2628 3.9970 1.7664
8 14.7579 0.0678 0.02907 34.3947 0.42907 2.3306 4.4713 1.9185
9 20.6610 0.0484 0.02034 49.1526 0.42034 2.3790 4.8585 2.0422
10 28.9255 0.0346 0.01432 69.8137 0.41432 2.4136 5.1696 2.1419
11 40.4957 0.0247 0.01013 98.7391 0.41013 2.4383 5.4166 2.2215
12 56.6939 0.0176 0.00718 139.2348 0.40718 2.4559 5.6106 2.2845
13 79.3715 0.0126 0.00510 195.9287 0.40510 2.4685 5.7618 2.3341
14 111.1201 0.0090 0.00363 275.3002 0.40363 2.4775 5.8788 2.3729
15 155.5681 0.0064 0.00259 386.4202 0.40259 2.4839 5.9688 2.4030
16 217.7953 0.0046 0.00185 541.9883 0.40185 2.4885 6.0376 2.4262
17 304.9135 0.0033 0.00132 759.7837 0.40132 2.4918 6.0901 2.4441
18 426.8789 0.0023 0.00094 1064.70 0.40094 2.4941 6.1299 2.4577
19 597.6304 0.0017 0.00067 1491.58 0.40067 2.4958 6.1601 2.4682
20 836.6826 0.0012 0.00048 2089.21 0.40048 2.4970 6.1828 2.4761
22 1639.90 0.0006 0.00024 4097.24 0.40024 2.4985 6.2127 2.4866
24 3214.20 0.0003 0.00012 8033.00 0.40012 2.4992 6.2294 2.4925
25 4499.88 0.0002 0.00009 11247 0.40009 2.4994 6.2347 2.4944
26 6299.83 0.0002 0.00006 15747 0.40006 2.4996 6.2387 2.4959
28 12348 0.0001 0.00003 30867 0.40003 2.4998 6.2438 2.4977
30 24201 0.00002 60501 0.40002 2.4999 6.2466 2.4988
32 47435 0.00001 0.40001 2.4999 6.2482 2.4993
34 92972 0.40000 2.5000 6.2490 2.4996
35 0.40000 2.5000 6.2493 2.4997

Compound Interest Factor Tables 609
50% TABLE 29 Discrete Cash Flow: Compound Interest Factors 50%
Single Payments Uniform Series Payments Arithmetic Gradients
Compound Present Sinking Compound Capital Present Gradient Gradient
Amount Worth Fund Amount Recovery Worth Present Worth Uniform Series
n FP PF AF FA AP PA PG AG
1 1.5000 0.6667 1.00000 1.0000 1.50000 0.6667
2 2.2500 0.4444 0.40000 2.5000 0.90000 1.1111 0.4444 0.4000
3 3.3750 0.2963 0.21053 4.7500 0.71053 1.4074 1.0370 0.7368
4 5.0625 0.1975 0.12308 8.1250 0.62308 1.6049 1.6296 1.0154
5 7.5938 0.1317 0.07583 13.1875 0.57583 1.7366 2.1564 1.2417
6 11.3906 0.0878 0.04812 20.7813 0.54812 1.8244 2.5953 1.4226
7 17.0859 0.0585 0.03108 32.1719 0.53108 1.8829 2.9465 1.5648
8 25.6289 0.0390 0.02030 49.2578 0.52030 1.9220 3.2196 1.6752
9 38.4434 0.0260 0.01335 74.8867 0.51335 1.9480 3.4277 1.7596
10 57.6650 0.0173 0.00882 113.3301 0.50882 1.9653 3.5838 1.8235
11 86.4976 0.0116 0.00585 170.9951 0.50585 1.9769 3.6994 1.8713
12 129.7463 0.0077 0.00388 257.4927 0.50388 1.9846 3.7842 1.9068
13 194.6195 0.0051 0.00258 387.2390 0.50258 1.9897 3.8459 1.9329
14 291.9293 0.0034 0.00172 581.8585 0.50172 1.9931 3.8904 1.9519
15 437.8939 0.0023 0.00114 873.7878 0.50114 1.9954 3.9224 1.9657
16 656.8408 0.0015 0.00076 1311.68 0.50076 1.9970 3.9452 1.9756
17 985.2613 0.0010 0.00051 1968.52 0.50051 1.9980 3.9614 1.9827
18 1477.89 0.0007 0.00034 2953.78 0.50034 1.9986 3.9729 1.9878
19 2216.84 0.0005 0.00023 4431.68 0.50023 1.9991 3.9811 1.9914
20 3325.26 0.0003 0.00015 6648.51 0.50015 1.9994 3.9868 1.9940
22 7481.83 0.0001 0.00007 14962 0.50007 1.9997 3.9936 1.9971
24 16834 0.0001 0.00003 33666 0.50003 1.9999 3.9969 1.9986
25 25251 0.00002 50500 0.50002 1.9999 3.9979 1.9990
26 37877 0.00001 75752 0.50001 1.9999 3.9985 1.9993
28 85223 0.00001 0.50001 2.0000 3.9993 1.9997
30 0.50000 2.0000 3.9997 1.9998
32 0.50000 2.0000 3.9998 1.9999
34 0.50000 2.0000 3.9999 2.0000
35 0.50000 2.0000 3.9999 2.0000

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Chapter 5
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Chapter 19
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INDEX
A
A , 13, 43
Absolute cell referencing, 9, 547, 550
Accelerated Cost Recovery System (ACRS), 422
Accelerated write-off, 416, 450–52
Accounting
ratios, 563–65
statements, 561–63
Acid-test ratio, 564
Acquisition phase, 161
Activity based costing (ABC), 401–03
Additive weight technique, 282–83
A/F factor, 46
After-tax
and alternative selection, 456–62
cash flow, 448–50
debt versus equity financing, 267
and depreciation, 415, 453
international, 468–70
and MARR, 269, 456
and present worth, 456
rate of return, 458–62
spreadsheet analysis, 457–58
and WACC, 271
After-tax replacement analysis, 462–65
A/G factor, 53. See also Gradient, arithmetic
Alternative depreciation system (ADS), 426–27
Allocation variance, 399
Alternatives
cost, 131
defined, 6
do-nothing, 130–31
independent, 130–31, 132, 157, 203, 207, 244, 247 ( see also
Capital budgeting)
infinite life, 138, 157–60
mutually exclusive, 130–31, 132, 155, 208, 238, 248
revenue, 131
selection, 6
service, 131
in simulation, 533–40
Amortization, 415
Annual interest rate
effective, 99–105
nominal, 99–105
Annual operating costs (AOC), 6, 153, 297–98
and estimation, 388
Annual Percentage Rate (APR), 97
Annual Percentage Yield (APY), 97
Annual worth
advantages, 151
after-tax analysis, 456–58
of annual operating costs, 297–98
and B/C analysis, 235, 238–42
and breakeven analysis, 345–48
and capital-recovery-plus-interest, 153–54
equivalent uniform, 151
evaluation by, 155–60
and EVA, 153, 465–68
and future worth, 151
and incremental rate of return, 213–14
of infinite-life projects, 157–60
and inflation, 151, 377–78
and present worth, 151
and rate of return, 175, 213–14
and replacement analysis, 302–06, 307–10, 462–65
spreadsheet solutions, 156, 159, 458
when to use, 262
AOC. See Annual operating costs
A/P factor, 43–45, 153
Arithmetic gradient. See Gradient, arithmetic
Assets. See also Book value; Depreciation; Life; Salvage value
in balance sheet, 561
capital recovery, 153–54
return on, 564
sunk cost, 295
Attributes
evaluating multiple, 282–83
identifying, 278–79
weighting, 279–81
Average. See Expected value
Average cost per unit, 345
Average tax rate, 447
B
Balance sheet, 267, 561–63
Base amount
defined, 50
and shifted gradients, 80–82
Basis, unadjusted, 416
B/C. See Benefit/cost ratio
Before-tax rate of return
and after-tax, 459
calculation, 173–90
Bell-shaped curve. See Normal distribution
Benefit and cost difference, 236
Benefit/cost ratio
calculation, 235–36
conventional, 235
incremental analysis, 238–39
modified, 235–36
for three or more alternatives, 242–46
for two alternatives, 238–42
when to use, 262
Benefits
direct versus implied, 242
in public projects, 231, 235, 242
(beta), 274
Bonds
and debt financing, 271–72
and inflation, 372, 385
interest computation, 190
payment periods, 190
present worth, 191–92
for public sector projects, 232
rate of return, 190–92
types, 191
Book depreciation, 415, 417, 466
Book value
declining balance method, 419–22
defined, 416
double declining balance method, 419–22
and EVA, 466

612 Index
Book value (continued)
MACRS method, 423
versus market value, 416
straight line method, 418
sum-of-years-digits method, 430, 557
unit-of-production method, 431
Borrowed money, 267
Borrowing rate, 185–87
Bottom-up approach, 388–89
Breakeven analysis. See also PW vs. i graph
average cost per unit, 345
fixed costs, 341
and Goal Seek, 353–54
and make-buy decisions, 341, 347
and payback, 351–54
and rate of return, 210–12, 460
versus sensitivity analysis, 341, 485
single project, 341–45
spreadsheet application, 212, 352–54
three or more alternatives, 348
two alternatives, 345–47
variable costs, 341
Breakeven point, 341, 489
Budgeting. See Capital budgeting
Bundles, 325, 327
Business ratios, 563–65
C
Canada, depreciation and taxes, 468
Capital
cost of ( see Cost of capital)
cost of invested, 466
debt versus equity, 27, 267
description, 3
limited, 268, 323
unrecovered, 295
Capital asset pricing model (CAPM), 274
Capital budgeting
description, 323–25
equal life projects, 325–26
linear programming, 329–31
mutually exclusive bundles, 325, 327
present-worth use, 325–29
reinvestment assumption, 324, 328–29
spreadsheet solution, 330–31
unequal life projects, 327–29
Capital financing. See also Cost of capital
debt, 27, 267, 271–73
equity, 27, 267, 273–75
mixed (debt and equity), 275–77
Capital gains
defined, 454
short-term and long-term, 454
taxes for, 454
Capital losses
defined, 454
taxes for, 454
Capital rationing. See Capital budgeting
Capital recovery, 153–54. See also A/P factor; Depreciation
defined, 153
and economic service life, 297–99
and EVA, 468
and inflation, 377–78
and replacement analysis, 306
Capital recovery factor, 43–45
and equivalent annual worth, 153
Capitalized cost
in alternative evaluation, 138–42
and annual worth, 138
and public projects, 244–46
CAPM. See Capital asset pricing model
Carry-back and carry-forward, 454
Case studies
after-tax analysis, 482–83
alternative description, 38
annual worth, 93, 171
breakeven analysis, 363–64
compound interest, 70–71
debt versus equity financing, 290–91, 482–83
economic service life, 321
energy, 36, 363–64
ethics, 412–13
house financing, 124–25
indirect costs, 411–12
inflation, 385
multiple attributes, 511–13
multiple interest rates, 200–01
public project, 259–60, 511–13
rate of return, 200–01, 385
replacement analysis, 321
sale of business, 227
sensitivity analysis, 226, 510, 544–45
simulation, 544–45
social security, 149
Cash flow
after tax, 448–50
before tax, 449–50
beyond study period, 133–34
continuous, 116
conventional series, 180
defined, 15
diagramming, 16–18
discounted, 129
estimating, 15, 231, 387–90
incremental, 203–06, 213
inflow and outflow, 15
net, 15
and payback period, 349–51
nonconventional, 180–90
and public sector projects, 231
recurring and nonrecurring, 139
and replacement analysis, 306
revenue versus cost, 131, 204
and rule of signs, 180–84, 219
and simulation, 533–38
using actual versus incremental, 213, 240, 460
zero, 78, 553, 555
Cash flow after taxes (CFAT), 448–50, 456–65
and EVA, 465–68
Cash flow before taxes (CFBT), 448–50, 473
Cash flow diagrams, 16–18
partitioned, 55
Cell references, spreadsheet, 29, 547, 550
CFAT. See Cash flow after taxes
CFBT. See Cash flow before taxes
Challenger
in B/C analysis, 242–46
in replacement analysis, 294, 302–12, 462–64
in ROR analysis, 214–19
in service sector projects, 248–49
China, depreciation and taxes, 468–69
Code of ethics, 7–8, 251, 404, 566–68
Common stock, 267, 273–75

Index 613
Compound amount factors
single payment (F/P), 40
uniform series (F/A), 46
Compound interest, 22, 24–25, 28. See also Compounding
Compounding
annual, 99–105
continuous, 114–16
and effective interest rate, 96
frequency, 97–98
interperiod, 113–14
period, 97–98
and simple interest, 21–23
Compounding period
continuous, 114–16
defined, 97
and effective annual rate, 102
and payment period, 106–14
Concepts, fundamental, summary, 573–76
Contingent projects, 323
Continuous compounding, 114–16
Contracts, types, 234
Conventional benefit/cost ratio, 235
Conventional cash flow series, 180
Conventional gradient, 51–57
Corporations
and capital, 4
financial worth, 466–68
leveraged, 275–77
Cost alternative, 131, 204, 216, 457, 460
Cost-effectiveness
analysis, 246–50
ratio, 246
Cost, life-cycle, 160–63
Cost-capacity equations, 394–95
Cost centers, 397, 401
Cost components, 387–88
Cost depletion, 427–29
Cost drivers, 401
Cost-estimating relationships, 394–97
Cost estimation
accuracy, 389
approaches, 388–89
cost-capacity method, 394–95
and cost indexes, 391–94
factor method, 395–97
and inflation, 12, 377
unit method, 390
Cost of capital
and debt-equity mix, 269–71
for debt financing, 271–73
defined, 26, 267
for equity financing, 273–75
versus MARR, 26, 267
weighted average, 27, 270, 275
Cost of goods sold, 397, 562–63, 565
Cost of invested capital, 466–67
Costs. See also Capital; Incremental cash flow; Opportunity cost
annual operating, 6, 153, 297–98
and annual worth, 153
of asset ownership, 153
direct, 387, 390–97
estimating, 388–97
EUAW( see Annual worth)
fixed, 341
indirect, 387, 396, 397–404
life-cycle, 160–63
marginal, 300–01
in public projects, 231, 235
sign convention, 15, 235
sunk, 295
variable, 341
Coupon rate, 190
Cumulative cash flow sign test, 181–84, 219
Cumulative distribution, 519–22
Current assets, 561
Current liabilities, 561–62
Current ratio, 563
D
DB function, 420, 550–51
DDB function, 420, 551
Debt capital, 267
Debt-equity mix, 269–71, 275–77
Debt financing, 269–71
on balance sheet, 267, 561
costs of, 271–73
and inflation, 374, 378
leveraging, 275–77
Debt ratio, 564
Decision making
attributes, 3–4, 278–83
under certainty, 517
and engineering economy role, 3–4
under risk, 517, 518–22
under uncertainty, 517
Decision trees, 494–98
Declining balance depreciation, 419–22
in Excel, 420, 550–51
Decreasing gradients, 51, 56, 83–86
Defender
in B/C analysis, 242–46
in replacement analysis, 294, 302–12, 462–64
in ROR analysis, 214–19
in service sector projects, 248–49
Deflation, 368–69
Delphi method, 278–79
Dependent projects, 323
Depletion
cost, 427–29
percentage, 427–29
Depreciation. See also Depreciation recapture; Rate of
depreciation; Replacement analysis
accelerated, 416, 419
ACRS, 422
alternative system, 426–27
and amortization, 415
basis, 416
book, 415, 417, 466
declining balance, 419–22, 550–51
defined, 415
double declining balance, 419–22, 551
and EVA, 466
general depreciation system (GDS), 426
half-year convention, 416, 424, 427
and income taxes, 415, 417, 445–68
MACRS, 417, 422–27
present worth of, 432–35
property class, 426
recovery period for, 416, 418, 426
rate of, 416
recovery rate, 416

614 Index
Depreciation (continued)
straight line, 418–19, 557
straight line alternative, 426–27
sum-of-years digits, 430, 557
switching methods, 432–38, 557–58
tax, 415, 417
unit-of-production, 431
Depreciation recapture
definition, 453
in replacement studies, 462–65
and taxes, 453, 461
Descartes’ rule, 181–84, 219
Design-build contracts, 234
Design stages, preliminary and detailed, 161, 163, 389
Design-to-cost approach, 388–89
Direct benefits, 242, 244
Direct costs, 387, 390–97
Disbenefits, 231, 235
Disbursements, 15, 177
Discount rate, 129, 232
Discounted cash flow, 129
Discounted payback analysis, 349
Discrete cash flows
compound interest factors (tables), 581–609
discrete versus continuous compounding, 114–16
and end-of-period convention, 15–16
Disposal phase, 161
Distribution. See also Probability distribution
defined, 519
normal, 520, 531–33
standard normal, 531–33
triangular, 520, 522
uniform, 520–21, 535, 538
Dividends
bonds, 190, 271–72
stocks, 273–75
Dominance, 248–49
Do-nothing alternative
and B/C analysis, 239, 242
defined, 6, 131
and independent projects, 132, 207, 244, 325–27
and present worth, 130–31
and rate of return, 207, 214, 216
Double declining balance, 419–22
in Excel, 420–22, 551
in switching, 432–38
and taxes, 451–52
Dumping, 368
E
Economic equivalence. See Equivalence
Economic service life (ESL), 294, 296–302
Economic value added, 153, 465–68
EFFECT function, 103, 551
Effective interest rate
annual, 99–100
for any time period, 105–06
of bonds, 191–92
and compounding periods, 100, 105
for continuous compounding, 114–16
defined, 96
and nominal rate, 96–97
Effective tax rate, 447, 462
Efficiency ratios, 563
End-of-period convention, 15–16
Engineering economy
Concepts, summary, 573–76
defined, 3
study approach, 4–7
terminology and symbols, 13
Equal service requirement, 131, 151, 213, 217, 240, 457
Equity financing, 26, 267
cost of, 273–75
Equivalence
calculations without tables, 569–72
compounding period greater than payment period, 112–14
compounding period less than payment period, 107–12
defined, 19–21
Equivalent uniform annual cost. See Annual worth
Equivalent uniform annual worth. See Annual worth
Error distribution. See Normal distribution
Estimation
and alternatives, 6
of cash flow, 6, 15–19
of costs, 388–97
and sensitivity analysis, 485, 490–91
before tax ROR, 459
Ethics
and cost estimation, 403–04
in public sector, 250–51
overview, 7–10
EUAC. See Annual worth
EUAW. See Annual worth
Evaluation criteria, 4, 262
Evaluation method, 261–64
Excel. See also Spreadsheet, usage in examples
absolute cell reference, 9, 547, 550
basics, 547–50
charts, 548–49
displaying functions, 30
embedding functions, 75, 156–57
error messages, 560
functions, in engineering economy, 28, 550–58
Goal Seek tool, 558–59
introduction, 27–30, 547–50
and linear programming, 330–31
random number generation, 556
Solver tool, 559–60
spreadsheet layout, 549–50
Expected value
computation, 492–94, 526–27, 530
and decisions under risk, 517
and decision trees, 497–98
defined, 492, 526
and real options, 501–02
in simulation, 538–39
Expenses, operating, 6, 153, 297–98
External rate of return, 185–90. See also Rate of return
F
F, 13
F/A factor, 46
F/G factor, 54
Face value, of bonds, 190
Factor method estimation, 395–97
Factors, compound interest
annual worth, 44, 46, 53
arithmetic gradient, 52–54

Index 615
capital recovery, 44
continuous compound interest, 114–16
derivations, 39–46, 52–54
equivalence without, 569–72
future worth, 40, 46, 54
geometric gradient, 58–59
interpolation, 48–50
multiple, 73–86
notation, 40, 44, 47, 53, 59
present worth, 40, 43, 52, 58
single payment, 39–42
sinking fund, 46
tables, 581–609
uniform series, 43–48, 53
Factory cost, 397, 562
Financial worth of corporations, 466, 561
First cost
and depreciation, 416
description, 6, 15
and estimation, 388–97
in replacement studies, 294–95, 462
Fiscal year, 561
Fixed assets, 561
Fixed costs, 341
Fixed income investment, 372, 385. See also Bonds
Fixed percentage method. See Declining balance depreciation
F/P factor, 40. See also Single payment factors
Future worth
and annual worth, 151
and effective interest rate, 100
evaluation by, 137
and inflation, 374–77
and multiple rates of return, 186–87
from present worth, 137
of shifted series, 73, 77–78
and spreadsheet solutions, 42, 48
when to use, 262
FV function, 28, 552
and shifted uniform series, 78–79
and single payments, 41
and uniform series, 46
G
Gains and losses, 454
Gaussian distribution. See Normal distribution
General depreciation system (GDS), 426
Geometric gradient
defined, 58
factors, 58–59
shifted, 82–86
Geometric series, and equivalence, 570–72
Goal Seek, 56–57, 86, 156–57, 189, 312, 354, 458, 558–59
Government projects. See Public sector projects
Gradient, arithmetic
base amount, 50
conventional, 51
decreasing, 51, 56
defined, 50
factors, 52–54
increasing, 51–55
shifted, 80–82, 83–86
spreadsheet use, 57
Graduated tax rates, 446–47, 448
Gross income, 445, 448
H
Half-year convention, 416, 424, 427
Highly leveraged corporations, 275–77
Hurdle rate. See Minimum attractive rate of return
Hyperinflation, 368, 377
I
IF function, 189, 245, 552
Implied benefits, 240, 242
Income
estimated annual, 6, 15
gross, 445, 448
net operating, 446, 459
taxable, 446, 454
Income statement, 562
basic equation, 562
ratios, 564
Income tax, 445–70
average tax rate, 447
and capital gains and losses, 454
and cash flow, 448–50
corporate, 447
and debt financing, 271–73
defined, 445
and depreciation, 450–56
effective rates, 447
international, 468–70
negative, 449
present worth of, 451–53
and rate of return, 458–62
rates, 446–48
and replacement studies, 462–65
state, 445, 447
tax savings, 449
and taxable income, 446
terminology, 445–46
Incremental benefit/cost analysis
for three or more alternatives, 242–46
for two alternatives, 238–42
Incremental cash flow
and benefit/cost analysis, 238–39
calculation, 203–06
and rate of return, 207–10, 213–14
Incremental rate of return
for multiple alternatives, 214–18
for two alternatives, 213–14
unequal lives, 207, 213
Independent projects
AW evaluation, 157
B/C evaluation, 244
and capital budgeting, 323–31
definition, 130
and do-nothing alternative, 130, 132, 157, 207, 244, 325, 327
and incremental cash flow, 207, 244
PW evaluation, 132
ROR evaluation, 207
service project evaluation, 247–48
Indexing, income taxes, 448
Indirect costs
and activity-based costing, 401–03
allocation variance, 399
in cost of goods sold statement, 562–63
defined, 398

616 Index
Indirect costs (continued)
and factor method, 396–97
rates, 398–99
Infinite life, 138–40, 157–60, 231, 244–45
Inflation
assumption, PW and AW analysis, 134, 151
and capital recovery, 377–78
constant-value, 367
definition, 12, 367
versus deflation, 368–69
and dumping, 368
and future worth, 374–77
high, 377
impact, 12, 367–68
and interest rates, 368
market adjusted, 368
and MARR, 368, 375–77
and present worth, 369–74
and sensitivity analysis, 485
Initial investment. See First cost
Installment financing, 175
Intangible factors, 6. See also Multiple attribute evaluation
Integer linear programming, 329–31
Interest. See also Interest rate(s)
compound, 22, 24, 29
continuous compounding, 114–16
defined, 10
interperiod, 112–14
rate, 10, 12
simple, 21, 24
Interest period, 10, 12
Interest rate(s). See also Effective interest rate; Rate of return
and breakeven analysis, 345
definition, 10–12
Excel functions, 61, 177, 553, 557
expressions, 98
inflation-adjusted, 368, 370
inflation free (real), 368, 374
interpolation, 61–63
market, 368, 370
multiple, 180–84
nominal versus effective, 96–99
for public sector, 232
and risk, 267, 274
and sensitivity analysis, 485
on unrecovered balance (ROR), 173–75
varying over time, 116–17
Interest tables
discrete compounding, 581–609
interpolation, 48–50
Internal rate of return. See also Rate of return
and annual worth, 175
definition, 173
versus external ROR, 185
and present worth, 175
spreadsheet solution, 177
International aspects
contracts, 234
corporate taxes, 468–70
cost estimation, 388–89
deflation, 368–69
depreciation, 416–17, 470–72
dumping, 368
inflation aspects, 368, 377
value-added tax, 470–72
Interperiod interest, 112–14
Interpolation, in interest rate tables, 48–50
Inventory turnover ratio, 564–65
Invested capital, cost of, 466–67
Investment(s). See also First cost
extra, 206–07, 214–18
fi xed income, 372, 385
net, 188
permanent, 138–41, 157–60
Investment rate, 185–90
IPMT function, 552
IRR function, 61, 63, 177–79, 217, 460, 553
L
Land, 416
Lang factors, 395
Least common multiple
and annual worth, 151–52
assumptions, 134
in evaluation methods, 262–63
and future worth, 137
and incremental cash flow, 204
and incremental rate of return, 207–09, 213–14, 218–19
and independent projects, 324, 329
and present worth, 133–37
in spreadsheet analysis, 205
versus study period, 134, 136
Leveraging, 275–77
Liabilities, 561
Life
finite, 141
and income taxes, 452–53
infinite or very long, 138–41, 157–58
minimum cost, 297
recovery period, 416
in simulation, 534–35, 538
unknown, 61, 63–64
useful, 6, 416
Life cycle, and annual worth, 160–62
Life-cycle costs, 160–63
Likert scale, 281
Linear programming, 329–31
Lives
equal, 131, 155, 204, 240
perpetual, 138, 157
unequal, 133, 155, 204, 207
Loan repayment, 24–25
M
MACRS (Modified Accelerated Cost Recovery System)
in CFAT example, 450
depreciation rates, 423–24
PW of, depreciation, 432
recovery period, 423, 426–27
spreadsheet function, 424, 558
straight line alternative (ADS), 426–27
switching, 432, 435–38
U.S., required, 417, 422
Maintenance and operating (M&O) costs, 6, 153. See also Annual
operating cost
Make-or-buy decisions, 341, 347. See also Breakeven analysis
Marginal costs, 300–01
Marginal tax rates, 446–48

Index 617
Market interest rate, 368
Market value
and depreciation, 416
in ESL analysis, 297, 300–01
estimating, 301–02
in replacement analysis, 294, 302–06
as salvage value, 153, 294
and study period, 134
MARR. See Minimum attractive rate of return
Mean. See Expected value
Mean squared deviation, 528
Measure of worth, 4, 6, 262
Median, 527
Mexico, depreciation and taxes, 469
Minimum attractive rate of return
after-tax, 269, 456–58
and bonds, 192
and capital budgeting, 323–27, 328–29
definition, 26, 267
establishing, 26–27, 267–69, 275
as hurdle rate, 26
inflation-adjusted, 368, 375–76
and rate of return, 203, 206, 208, 213, 214–18
and reinvestment, 324, 328–29
in sensitivity analysis, 485–87
before tax, 269
and WACC, 27, 267, 275
Minimum cost life. See Economic service life
MIRR function, 186, 553–54
M&O costs. See Annual operating costs; Maintenance and
operating costs
Mode, 522, 527
Modified benefit/cost ratio, 235–36
Modified ROR approach, 185–87
Monte Carlo simulation, 533–39
Most likely estimate, 490–91
Multiple alternatives
breakeven analysis, 348
incremental benefit/cost analysis, 242–46
incremental rate of return, 214–18
Multiple attribute evaluation, 278–83
Multiple rate of return
definition, 179–80
determining, 181–84
presence of, 180–81
removing, 184–90
Municipal bonds, 190–92
Mutually exclusive alternatives, 130–31
and annual worth, 155–57
and B/C, 238–44
evaluation method selection, 261–64
and present worth, 132–37
and rate of return, 208–10
and service projects, 247–49
N
Natural resources. See Depletion
Net cash flow, 15
Net operating income (NOI), 446, 459
Net investment procedure, 187–90
Net operating profit after taxes (NOPAT), 446, 466
Net present value. See NPV function; Present worth
Net profit after taxes (NPAT), 446
Net worth, 561
NOMINAL function, 103, 554
Nominal interest rate
annual, 96, 103
of bonds, 190–92
definition, 96
and effective rates, 96–98
Nonconventional cash flow series, 180–84
Noneconomic factors, 6
Nonowner’s viewpoint, 295
Nonrecurring cash flows, 139
Nonsimple cash flow series, 180–84
No-return (simple) payback, 349–50
Normal distribution, 520, 531–33
Norstrom’s criterion, 181–84, 219, 460
Notation for factors, 40, 44, 47, 53, 59
NPER function, 554–55
and payback, 350, 353
and unknown n, 61, 63–64
NPV function, 555
for arithmetic gradient, 57, 86
embedding in PMT, 75, 156
geometric gradients 86
independent projects, 328
and present worth, 135–36
in PW vs. i graphs, 178–79
sensitivity analysis, 486–87, 489
and shifted series, 75, 86
NSPE (National Society of Professional Engineers), 7,
404, 566–68
O
Obsolescence, 294
One-additional-year replacement study, 302–05
Operating costs. See Annual operating costs
Operations phase, 161, 163
Opportunity cost, 27, 267, 324
and replacement analysis, 306
Optimistic estimate, 490–91
Order of magnitude, 389
Overhead rates. See Indirect costs
Owner’s equity, 267, 561
P
P, 13, 40
P/A factor,43, 53. See also Geometric gradient;
Uniform series
Payback analysis
and breakeven analysis, 351–52
calculation, 349–51
definition, 348–49
limitations, 349
spreadsheet analysis, 350–54
Payment period
of bonds, 190
defined, 106
equals compounding period, 106, 107, 109–11
longer than compounding period, 109
shorter than compounding period, 112–14
single amount, 107–09
Payout period. See Payback analysis
Percentage depletion. 427–29
Permanent investments, 138–41, 157–58
Personal property, 416, 423, 426

618 Index
Perspective
for public sector analysis, 232–34
for replacement analysis, 295
Pessimistic estimate, 490–91
P/F factor, 40
P/G factor, 53. See also Gradient, arithmetic
Phaseout phase, 161
Planning horizon. See Study period
PMT function, 28, 555
and after-tax analysis, 457
and annual worth, 44, 155–57
and arithmetic gradient, 54
and capital recovery, 154
and economic service life, 298–99
and embedded NPV, 75, 80, 156
and geometric gradient, 59
and random single amounts, 79–80
and shifted series, 75–76
and sinking fund factor, 44
and uniform series, 75, 79
Point estimates, 15, 517
Power law and sizing model, 394–95
PPMT function, 555–56
Preferred stock, 267, 273
Present value. See Present worth
Present worth
after-tax analysis, 456–58
and annual worth, 151
assumptions, 134
and B/C analysis, 235
of bonds, 191–92
and breakeven analysis, 345–46
and capital budgeting, 325–29
of depreciation, 432
for equal lives, 132–33
evaluation method, 261–62
geometric gradient series, 82–86
income taxes, 451–53
and independent projects, 325–29
index, 332
and inflation, 369–74
and multiple interest rates, 181–84
and profitability index, 237
and rate of return, 175–76, 182–84, 458–62
and sensitivity analysis, 486–90
in shifted series, 73, 76, 80–86
in simulation, 534–39
single-payment factor, 40–41
for unequal lives, 133–37
Present worth factors
gradient, 50–53, 58–60
single payment factor, 40–41
uniform series, 43–45
Probability
in decision trees, 495–98
defined, 492, 518
and expected value, 492–94, 526–27
and standard deviation, 528
Probability distribution
of continuous variables, 519–22
defined, 519
of discrete variables, 519–20
properties, 526–28, 530
and samples, 523–26
in simulation, 533–39
Probability node, 495
Productive hour rate, 399
Profitability index, 237, 332
Profitability ratios, 563
Profit-and-loss statement, 562
Project net-investment, 187–90
Property class, 426
Property of independent random variables, 533
Public-private partnerships, 234–35
Public sector projects, 230–35
and annual worth, 157–58
B/C analysis, 235–38
capitalized cost, 138–42
characteristics, 231–32
design-build contracts, 234–35
profitability index, 237
public-private partnerships, 234–35
Purchasing power, 367, 374, 376
PV function, 28, 44, 556
versus NPV function, 556
and present worth, 41, 44
and single payment, 41
and uniform series present worth, 44
PW vs. i graph, 177–79, 182–84, 210–13, 460–62
R
RAND function, 523–25, 556
Random numbers, 523–25, 556
generation, 538, 556
Random samples, 523–26, 535–39
Random variable
continuous, 519–20, 525
cumulative distribution, 519–22, 525–26
defined, 518
discrete, 519–20, 523
expected value, 526, 528, 530
probability distribution of, 519
standard deviation, 527–28, 530
Range, 15, 530
Rank and rate technique, 282
Ranking inconsistency, 213, 216, 462
RATE function, 61–62, 177, 557
Rate of depreciation
declining balance, 419
defined, 416
MACRS, 422–23, 435–38
straight line, 418, 420
sum-of-years digits, 430
Rate of return. See also Incremental rate of return
after-tax, 458–62
and annual worth, 175, 213–14, 218
of bonds, 190–92, 272
breakeven, 210–12, 460
in capital budgeting, 332–34
cautions, 179–80
on debt capital, 272–73, 276
defined, 12, 173, 175
on equity capital, 273–74, 276
evaluation method, 261–62
external, 185–90
on extra investment, 206
incremental, 206, 207–10
and independent projects, 207
and inflation, 12–13, 368, 374–75

Index 619
installment financing, 175
internal, 173, 175, 185
minimum attractive ( see Minimum attractive rate of return)
modified ROR approach, 185–87
multiple, 180–90
and mutually exclusive alternatives, 206, 207–18
and present worth, 175, 177–79, 207–09, 218
ranking inconsistency, 213, 216, 462
return on invested capital (ROIC) approach, 185, 187–90
spreadsheet solution, 177, 179, 211–13, 216–19
Ratios, accounting, 563–65
Real interest rate, 368, 370, 374–75
Real options, 498–503
and decision trees, 500
definition, 499
Real property, 416, 423–24, 426
Recovery period
defined, 416
effect on taxes, 452–53
MACRS, 423–24, 426–27
straight line option, 426
Recovery rate. See Rate of depreciation
Recurring cash flows, 139
Reinvestment, assumption in capital budgeting,
324, 328–29
Reinvestment rate. See Investment rate
Repayment of loans, 24–25
Replacement analysis, 292–313, 462–65
after-tax, 462–65
annual worth, 295, 302–06
and capital losses, 454, 462
cash flow approach, 306
depreciation recapture, 462–64
and economic service life, 296–99, 305
first costs, 294–96
and marginal costs, 300–01
market value, 294, 301
need for, 294
one-additional year, 302–05
opportunity cost approach, 306
overview, 302
and study periods, 307–12
sunk costs, 295
terminology, 294
viewpoint, 295
Replacement life. See Economic service life
Replacement value, 312
Retained earnings, 27, 267, 273
Retirement life. See Economic service life
Return on assets ratio, 564
Return on invested capital, 187–90
Return on investment (ROI), 12, 173. See also Rate of return
Return on sales ratio, 564
Revenue alternatives, 131, 204, 214, 237, 460
Risk
and debt-equity mix, 275–77
and decision making, 517, 526–30, 533–39
and decision trees, 494–98
description, 515
and MARR, 267–69
and payback analysis, 349
and random sampling, 523
and real options, 498–502
Risk-free investment, 26, 274
ROI. See Return on investment
Root mean square deviation, 528
ROR. See Rate of return
Rule of signs, 181
S
Safe investment, 26, 190, 274
Sales, return on, 564
Salvage value. See also Market value
and capital recovery, 153, 297
defined, 6, 153
and depreciation, 416, 418, 420, 423, 430
and market value, 294, 297–98
and public projects, 235
in PW analysis, 132, 134–37
in replacement analysis, 294, 297, 303–06, 464
and trade-in value, 294, 464
Sampling, 523–26
Savings, tax, 449, 463–65
Scatter charts. See xy Excel charts
Screening projects, 349, 351
Section 179 deduction, 417
Section 1231 transactions, 454
Security, defined, 274
Sensitivity analysis. See also Breakeven analysis
description, 485
and Excel cell referencing, 29, 547
of one parameter, 485–87
spider graph, 488
with three estimates, 490–91
two alternatives, 488–90
Service alternative. See Cost alternative
Service sector projects
analysis, 246–50
definition, 246
dominance, 248–49
Shifted gradients, 80–86
Shifted series, 73–80
Sign changes, number of, 181–84
Simple cash flow series, 180
Simple interest, 21–23
Simulation, Monte Carlo, 517, 533–40
Single payment compound amount (F/P) factor, 40
Single payment factors, 39–42
Single payment present worth (P/F) factor, 40
Sinking fund (A/F) factor, 46
SLN function, 418, 557
Social discount rate, 232
Solvency ratios, 563
Solver, 330–31, 559–60
Spreadsheet, usage in examples. See also Excel
annual worth, 155–56, 159, 218, 299, 301, 304,
305, 353, 458, 467
B/C analysis, 245
breakeven analysis, 352, 353–54
cash flow after tax (CFAT), 450, 456, 458, 461, 462
compound interest, 29–30
depreciation, 422, 425, 434
EVA, 467
and factor values, 49
independent projects, 328, 331
inflation, 372
layout, 549–50
multiple attributes, 283
nominal and effective interest, 103–04

620 Index
Spreadsheet (continued)
present worth, 136, 209, 212, 218, 311, 354, 458,
461, 489, 501, 539
rate of return, 63, 182, 184, 189, 192, 209, 212,
217, 218, 461, 462, 501
replacement analysis, 299, 301, 304, 305, 311, 465
replacement value, 312
sensitivity analysis, 487, 489
simulation, 538–39
Staged funding, 494–503
Standard deviation
for continuous variable, 530
definition, 527–28
for discrete variable, 528–29
Standard normal distribution, 531–33
Stocks
CAPM model, 274
common, 267, 274
in equity financing, 273–75, 276
preferred, 267, 273
Straight line alternative, in MACRS, 426–27
Straight line depreciation, 418–19
Straight line rate, 418
Study period
and AW evaluation, 155
and equal service, 133
and FW analysis, 137
and PW evaluation, 133–34
and replacement analysis, 302, 307–11
and salvage value, 153
spreadsheet example, 136, 310–11
Sum-of-years digits depreciation, 430
Sunk costs, 295
SYD function, 430, 557
System, phases of, 160–61
T
Tax depreciation, 415–16, 417
Taxable income, 446–70
and CFAT, 448–50, 456
and depreciation, 446, 450–53
negative, 449, 463
and taxes, 446–48, 463–65
Taxes. See After-tax; Income tax; Taxable income
Time, 13
Time value of money
defined, 4
and equivalence, 19
factors to account for, 39–61
and no-return payback, 349–50
Total cost relation, 342–45. See also Breakeven analysis
Trade-in value, 6, 294, 416. See also Market value; Salvage value
Treasury securities, 26, 190
Triangular distribution, 520, 522
U
Unadjusted basis, 416
Uncertainty, 515, 517
Uniform distribution, 520–21, 535, 538
Uniform gradient. See Gradient, arithmetic
Uniform percentage method. See Declining balance depreciation
Uniform series
compound amount (F/A) factor, 46
compounding period greater than payment period, 112–14
compounding period less than payment period, 109–12
description, 13
present worth (P/A) factor, 43
shifted, 73–80
Unit method, 390–91
Unit-of-production depreciation, 431
Unknown interest rate, 61–63
Unknown years (life), 61, 63–64
Unrecovered balance, 173–74
V
Value, resale, 6. See also Salvage value; Trade-in value
Value added analysis, after tax. See Economic value added
Value-added tax, 470–72
Variable. See Random variable
Variable costs, 341
Variance
in cost allocation, 399
description, 528
formula for, 528, 530
and normal distribution, 531
VDB function, 424, 433–34, 557–58
W
WACC. See Weighted average cost of capital
Websites, 580
Weighted attribute method, 282–83
Weighted average cost of capital, 27, 270–71, 275–77
Working capital, 563
Worth, measures of, 4, 6, 129, 534
X
xy Excel charts, 182, 184, 212, 352, 461, 487, 548–49
Y
Year(s)
and end-of-period convention, 15–16
fiscal versus calendar, 561
half-year convention, 416, 424, 427
symbols, 13
unknown, 61, 63–64

Format for Spreadsheet Functions on Excel ©
Present worth: Contents of ( )
PV(i%, n, A, F)
NPV(i%,second_cell:last_cell) first_cell
Future worth:
FV(i%, n, A, P)
Annual worth:
PMT(i%, n, P, F)
PMT(i%, n, NPV)
Number of periods (years):
NPER(i%, A, P, F)
for constant A series; single F value
for varying cash flow series
for constant A series; single P value
for single amounts with no A series
to find AW from NPV; embed NPV function
for constant A series; single P and F
(Note: The PV, FV, and PMT functions change the sense of the sign. Place a minus in front of the
function to retain the same sign.)
Rate of return:
RATE(n, A, P, F)
IRR(first_cell:last_cell)
Interest rate:
EFFECT(r%, m)
NOMINAL(i%, m)
Depreciation:
SLN(P, S, n)
DDB(P, S, n, t, d)
DB(P,S,n,t)
for constant A series; single P and F
for varying cash flow series
for nominal r, compounded m times per period
for effective annual i, compounded m times
per year
straight line depreciation for each period
double declining balance depreciation for
period t at rate d (optional)
declining balance, rate determined by the
function
Relations for Discrete Cash Flows with End-of-Period Compounding
Factor Notation
Sample Cash Flow
Type Find/Given and Formula Relation Diagram
Single
Amount
Uniform
Series
Arithmetic
Gradient
Geometric
Gradient
FP
Compound
amount
PF
Present
worth
PA
Present
worth
AP
Capital
recovery
FA
Compound
amount
AF
Sinking
fund
P G G
Present
worth
A G G
Uniform
series
P g A 1 and g
Present
worth
(FP,i,n) (1 i) n
(PF,i,n)
1
(1 i)
n
F P(FP,i,n)
P F(PF,i,n)
(Sec. 2.1)
(PA,i,n) (1 i)n 1
i(1 i)
n
P A(PA,i,n)
(AP,i,n)
i(1 i) n
(1 i) n 1
(FA,i,n) (1 i)n 1
i
(AF,i,n)
i
(1 i)
n
1
A P(AP,i,n)
(Sec. 2.2)
F A(FA,i,n)
A F(AF,i,n)
(Sec. 2.3)
(PG,i,n) (1 i)n in 1
i 2 (1 i) n P G G(PG,i,n)
(AG,i,n) 1 n
i (1 i)
n
1
A G G(AG,i,n)
(Gradient only) (Sec. 2.5)
A 1 [ 1 ( 1 g n
———
1 i )
———————
P g i g
A
n
1 ———
1 i
]
g i
g i
(Gradient and base A 1 ) (Sec. 2.6)
P G
0
P
0
P
1 2 n–1
…
A
A
1 2
…
…
0 1 2 n–1
…
A
A … A
…
0 1 2 3
G
2G
P g
A
n–1
F
A
F
n
A
n
n
A G A G A G
… A G A G
…
n
(n–1) G
A 1
A 1 (1+g) A 1 (1+g)n–1
0 1 2
…
…
n–1
n
VBD(P,0, n,MAX(0, t1.5),
MIN(n, t0.5), d)
MACRS depreciation for year t at rate d for
DDB or DB method
Logical IF function:
IF(logical_test,value_if_true,value_if_false)
for logical two-branch operations